Convergence of $sum_{n=1}^inftyleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)$ where $a_1=1$ and $a_n=2-frac...












2












$begingroup$


Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
converges to?




I started by expanding the sum:



$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



$=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



$=frac{1}{a_1^2}$



$=1$




Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






sequences-and-series convergence







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
    converges to?




    I started by expanding the sum:



    $sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



    $=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



    $=frac{1}{a_1^2}$



    $=1$




    Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



    PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






    sequences-and-series convergence







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
      converges to?




      I started by expanding the sum:



      $sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}$



      $=1$




      Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



      PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






      sequences-and-series convergence







      share|cite|improve this question











      $endgroup$




      Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
      converges to?




      I started by expanding the sum:



      $sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}$



      $=1$




      Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



      PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






      sequences-and-series convergence




      sequences-and-series convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 22 hours ago









      Asaf Karagila♦

      303k32429761




      303k32429761










      asked yesterday









      s0ulr3aper07s0ulr3aper07

      3119




      3119






















          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$

          Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
          Since $a_1=1$ and $a_nto 2$, we have
          $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$


            Infinite sums can behave in strange ways




            True!




            Sometimes more than one solution may follow logically




            Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



            To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
            $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
            &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
            &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
            See if you can finish the job from here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
              $endgroup$
              – s0ulr3aper07
              yesterday






            • 2




              $begingroup$
              Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
              $endgroup$
              – David
              yesterday





















            3












            $begingroup$

            $a_n$ does not approach zero,
            so the end term can not
            be disregarded.



            In other words,
            the sum up to $n$ is
            $frac1{a_1^2}
            -frac1{a_n^2}
            to 1-frac14
            =frac34
            $            .






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3099389%2fconvergence-of-sum-n-1-infty-left-frac1a-n2-frac1a-n12-right%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              7












              $begingroup$

              Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
              Since $a_1=1$ and $a_nto 2$, we have
              $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
                Since $a_1=1$ and $a_nto 2$, we have
                $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
                  Since $a_1=1$ and $a_nto 2$, we have
                  $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
                  Since $a_1=1$ and $a_nto 2$, we have
                  $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Eclipse SunEclipse Sun

                  7,4591437




                  7,4591437























                      4












                      $begingroup$


                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday


















                      4












                      $begingroup$


                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday
















                      4












                      4








                      4





                      $begingroup$


                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.






                      share|cite|improve this answer









                      $endgroup$




                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered yesterday









                      DavidDavid

                      68.2k664126




                      68.2k664126












                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday




















                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday


















                      $begingroup$
                      I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                      $endgroup$
                      – s0ulr3aper07
                      yesterday




                      $begingroup$
                      I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                      $endgroup$
                      – s0ulr3aper07
                      yesterday




                      2




                      2




                      $begingroup$
                      Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                      $endgroup$
                      – David
                      yesterday






                      $begingroup$
                      Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                      $endgroup$
                      – David
                      yesterday













                      3












                      $begingroup$

                      $a_n$ does not approach zero,
                      so the end term can not
                      be disregarded.



                      In other words,
                      the sum up to $n$ is
                      $frac1{a_1^2}
                      -frac1{a_n^2}
                      to 1-frac14
                      =frac34
                      $                      .






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        $a_n$ does not approach zero,
                        so the end term can not
                        be disregarded.



                        In other words,
                        the sum up to $n$ is
                        $frac1{a_1^2}
                        -frac1{a_n^2}
                        to 1-frac14
                        =frac34
                        $                        .






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          $a_n$ does not approach zero,
                          so the end term can not
                          be disregarded.



                          In other words,
                          the sum up to $n$ is
                          $frac1{a_1^2}
                          -frac1{a_n^2}
                          to 1-frac14
                          =frac34
                          $                          .






                          share|cite|improve this answer









                          $endgroup$



                          $a_n$ does not approach zero,
                          so the end term can not
                          be disregarded.



                          In other words,
                          the sum up to $n$ is
                          $frac1{a_1^2}
                          -frac1{a_n^2}
                          to 1-frac14
                          =frac34
                          $                          .







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          marty cohenmarty cohen

                          73.4k549128




                          73.4k549128






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3099389%2fconvergence-of-sum-n-1-infty-left-frac1a-n2-frac1a-n12-right%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness. ...
                              Read more

                              Alcedinidae

                              Image
                              Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
                              Read more

                              Origin of the phrase “under your belt”?

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
                              Read more