Is the sequence of sums of inverse of natural numbers bounded? [duplicate]
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This question already has an answer here:
Why does the series $sum_{n=1}^inftyfrac1n$ not converge?
21 answers
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?
calculus sequences-and-series harmonic-numbers
asked 2 days ago
kyle campbellkyle campbell
745
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marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 8 more comments
$begingroup$
This question already has an answer here:
Why does the series $sum_{n=1}^inftyfrac1n$ not converge?
21 answers
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?
calculus sequences-and-series harmonic-numbers
asked 2 days ago
kyle campbellkyle campbell
745
$endgroup$
marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
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– Lord Shark the Unknown
2 days ago
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@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
2 days ago
1
$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
2 days ago
1
$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
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– Kavi Rama Murthy
2 days ago
1
$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
2 days ago
|
show 8 more comments
$begingroup$
This question already has an answer here:
Why does the series $sum_{n=1}^inftyfrac1n$ not converge?
21 answers
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?
calculus sequences-and-series harmonic-numbers
asked 2 days ago
kyle campbellkyle campbell
745
$endgroup$
This question already has an answer here:
Why does the series $sum_{n=1}^inftyfrac1n$ not converge?
21 answers
I'm reading through Spivak Ch.22 (Infinite Sequences) right now. He mentioned in the written portion that it's often not a trivial matter to determine the boundedness of sequences. With that in mind, he gave us a sequence to chew on before we learn more about boundedness. That sequence is:
$$1, 1+frac{1}{2}, 1+frac{1}{2}+frac{1}{3}, 1+frac{1}{2}+frac{1}{3}+frac{1}{4}, . . .$$
I know that a sequence is bounded above if there is a number $M$ such that $a_nleq M$ for all $n$. Any hints here?
This question already has an answer here:
Why does the series $sum_{n=1}^inftyfrac1n$ not converge?
21 answers
calculus sequences-and-series harmonic-numbers
calculus sequences-and-series harmonic-numbers
asked 2 days ago
kyle campbellkyle campbell
745
asked 2 days ago
kyle campbellkyle campbell
745
edited yesterday
kyle campbell
asked 2 days ago
kyle campbellkyle campbell
745
asked 2 days ago
kyle campbellkyle campbell
745
asked 2 days ago
kyle campbellkyle campbell
745
745
marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud D., Martin R, user159517, Saad, Matthew Towers 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
2 days ago
1
$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
2 days ago
1
$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
2 days ago
|
show 8 more comments
3
$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
2 days ago
1
$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
2 days ago
1
$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
2 days ago
3
3
$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
2 days ago
$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
2 days ago
1
1
$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
2 days ago
1
1
$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
1
$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
2 days ago
$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
2 days ago
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
answered 2 days ago
Tyler6Tyler6
834313
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1
$begingroup$
Indeed! Thank you for that input.
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– kyle campbell
2 days ago
3
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
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– A. Arredondo
2 days ago
add a comment |
$begingroup$
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
answered 2 days ago
MustangMustang
3507
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
answered 2 days ago
Tyler6Tyler6
834313
$endgroup$
1
$begingroup$
Indeed! Thank you for that input.
$endgroup$
– kyle campbell
2 days ago
3
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
$endgroup$
– A. Arredondo
2 days ago
add a comment |
$begingroup$
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
answered 2 days ago
Tyler6Tyler6
834313
$endgroup$
1
$begingroup$
Indeed! Thank you for that input.
$endgroup$
– kyle campbell
2 days ago
3
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
$endgroup$
– A. Arredondo
2 days ago
add a comment |
$begingroup$
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
answered 2 days ago
Tyler6Tyler6
834313
$endgroup$
This is a classic example of how our intuition can be wrong concerning infinities. As @Kavi said in the comments, this is the harmonic series and indeed diverges. I believe that Spivak provided this as an example to “chew on” in order to show how non trivial boundedness can be, as on first glance many people believe this sequence should be bounded.
answered 2 days ago
Tyler6Tyler6
834313
answered 2 days ago
Tyler6Tyler6
834313
answered 2 days ago
Tyler6Tyler6
834313
answered 2 days ago
Tyler6Tyler6
834313
834313
1
$begingroup$
Indeed! Thank you for that input.
$endgroup$
– kyle campbell
2 days ago
3
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
$endgroup$
– A. Arredondo
2 days ago
add a comment |
1
$begingroup$
Indeed! Thank you for that input.
$endgroup$
– kyle campbell
2 days ago
3
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
$endgroup$
– A. Arredondo
2 days ago
1
1
$begingroup$
Indeed! Thank you for that input.
$endgroup$
– kyle campbell
2 days ago
$begingroup$
Indeed! Thank you for that input.
$endgroup$
– kyle campbell
2 days ago
3
3
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
$endgroup$
– A. Arredondo
2 days ago
$begingroup$
This is not the harmonic series, it is the sequence of partial sums of the harmonic series. A sequence and a series are two different things. Other important sequences linked to a series are the sequence of the terms (${1/n}$ in this case) and the sequence of partial rests.
$endgroup$
– A. Arredondo
2 days ago
add a comment |
$begingroup$
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
answered 2 days ago
MustangMustang
3507
$endgroup$
add a comment |
$begingroup$
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
answered 2 days ago
MustangMustang
3507
$endgroup$
add a comment |
$begingroup$
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
answered 2 days ago
MustangMustang
3507
$endgroup$
Indeed, you can notice that
$$a_1=1,a_4>2(=1+frac{1}{2}+frac{1}{4}+frac{1}{4}), a_8>frac{5}{2}(=a_4+frac{1}{8}+frac{1}{8}+frac{1}{8}+frac{1}{8}),a_{16}>3(=a_{8}+frac{1}{16}+frac{1}{16}+..8times)$$, so the general pattern is $a_{2^{2n}}>n+1$, so given an upper bound M, you can find a natural number n s.t. $ngeq M$ by archimedian property and so $a_{2^{2n}}>n+1>M$ and hence it can't be bounded.
answered 2 days ago
MustangMustang
3507
answered 2 days ago
MustangMustang
3507
answered 2 days ago
MustangMustang
3507
answered 2 days ago
MustangMustang
3507
3507
add a comment |
add a comment |
3
$begingroup$
It is good that you are attending to the definition of boundedness. So, can you write down a suitable $M$ with $a_nle M$ for all $n$?
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
@LordSharktheUnknown I'm not too sure what that $M$ would be. Can you give a hint?
$endgroup$
– kyle campbell
2 days ago
1
$begingroup$
Those aren't subsequences. In any case, are you sure you have transcribed Spivak's sequence correctly?
$endgroup$
– Lord Shark the Unknown
2 days ago
1
$begingroup$
It is just divergence of harmonic series. You can search the net for 'divergence of harmonic series'.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
A classic estimate: for each $n$, consider estimating $a_{2^n}$.
$endgroup$
– xbh
2 days ago