Calculate $sum_{n=1}^infty frac{n+1}{n(n+2)^2}$ using Basel Problem sum












3












$begingroup$


I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.



Any tips of how to proceed from here?






sequences-and-series summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24
















3












$begingroup$


I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.



Any tips of how to proceed from here?






sequences-and-series summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24














3












3








3





$begingroup$


I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.



Any tips of how to proceed from here?






sequences-and-series summation







share|cite|improve this question











$endgroup$




I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.



Any tips of how to proceed from here?






sequences-and-series summation




sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 31 at 11:12









YuiTo Cheng

2,40641037




2,40641037










asked Mar 31 at 9:10









Avi PAvi P

396




396












  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24


















  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24
















$begingroup$
Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
$endgroup$
– Dr. Mathva
Mar 31 at 9:24




$begingroup$
Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
$endgroup$
– Dr. Mathva
Mar 31 at 9:24










1 Answer
1






active

oldest

votes


















5












$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32












Your Answer








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%2f3169193%2fcalculate-sum-n-1-infty-fracn1nn22-using-basel-problem-sum%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32
















5












$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32














5












5








5





$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?






share|cite|improve this answer











$endgroup$



Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 31 at 18:54

























answered Mar 31 at 9:30









SilSil

5,65721745




5,65721745








  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32














  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32








1




1




$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
Mar 31 at 9:32




$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
Mar 31 at 9:32


















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%2f3169193%2fcalculate-sum-n-1-infty-fracn1nn22-using-basel-problem-sum%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