Decreasing sequence of positive numbers and convergent series (1 Viewer)

shongaponga · Apr 12, 2015

Hey guys, I am attempting to prove a question that I found from a problem set from Stanford Uni on series and sequences, however I'm having a little bit of difficulty. I've been at it for a while and so finally i've resorted to looking at the provided solution. The question and solution are provided below:

$\\ \text{Prove that if}~ \{{a_n}\} ~\text{is a decreasing sequence of positive numbers and}~ \sum_{n=1}^{\infty}{a_n} ~\text{converges, then} \lim_{n\rightarrow \infty} n{a_n}=0.$

Provided Solution:

$\\ ~\text{Let}~ \epsilon > 0. ~\text{Let}~S_n = \sum_{k=1}^{n}{a_k}. ~\text{Since}~ S_n $ converges$, S_n $ is Cauchy. Hence there exists some $N $ such that $ n, m \geq N $ \implies $ |S_n - S_m| < \epsilon. $ In particular for $ n \geq N $ we have $ (n-N)a_n \leq a_{N+1} + ... + a_n = |S_n - S_N| < \epsilon. $ Hence, $ \lim_{n\rightarrow \infty} (n-N)a_n=0. $ Thus we have$ \lim_{n\rightarrow \infty} N{a_n}=N\lim_{n\rightarrow \infty}{a_n}=0 $ too$, $ and this implies$ \lim_{n\rightarrow \infty} n{a_n} = \lim_{n\rightarrow \infty} (n-N)a_n + \lim_{n\rightarrow \infty}Na_n=0. ~~~~~~~~~~~$QED $$

Now i recognized that S_n was Cauchy and hence for $n, m \geq N $ \implies $ |S_n - S_m| < \epsilon,$ however this is as far as i was able to get (i had a bit more working but none of it got me much further towards a solution).

My questions are: (1) how is $(n-N)a_n \leq a_{N+1} + ... + a_n ?$ ie i simply can't wrap my head around how this inequality holds (i understand it must have something to do with the fact that a_n is decreasing) - if someone could explain this that would be great! And (2) how would one be able to recognize that (n-N)a_n <= |S_n - S_N| < epsilon would lead to the final result? (eg clues/hints from the question?) It just doesn't come to me intuitively.

Any help, (and alternative solutions) are much appreciated

InteGrand · Apr 12, 2015

shongaponga said:
Hey guys, I am attempting to prove a question that I found from a problem set from Stanford Uni on series and sequences, however I'm having a little bit of difficulty. I've been at it for a while and so finally i've resorted to looking at the provided solution. The question and solution are provided below:

$\\ \text{Prove that if}~ \{{a_n}\} ~\text{is a decreasing sequence of positive numbers and}~ \sum_{n=1}^{\infty}{a_n} ~\text{converges, then} \lim_{n\rightarrow \infty} n{a_n}=0.$

Provided Solution:

$\\ ~\text{Let}~ \epsilon > 0. ~\text{Let}~S_n = \sum_{k=1}^{n}{a_k}. ~\text{Since}~ S_n $ converges$, S_n $ is Cauchy. Hence there exists some $N $ such that $ n, m \geq N $ \implies $ |S_n - S_m| < \epsilon. $ In particular for $ n \geq N $ we have $ (n-N)a_n \leq a_{N+1} + ... + a_n = |S_n - S_N| < \epsilon. $ Hence, $ \lim_{n\rightarrow \infty} (n-N)a_n=0. $ Thus we have$ \lim_{n\rightarrow \infty} N{a_n}=N\lim_{n\rightarrow \infty}{a_n}=0 $ too$, $ and this implies$ \lim_{n\rightarrow \infty} n{a_n} = \lim_{n\rightarrow \infty} (n-N)a_n + \lim_{n\rightarrow \infty}Na_n=0. ~~~~~~~~~~~$QED $$

Now i recognized that S_n was Cauchy and hence for $n, m \geq N $ \implies $ |S_n - S_m| < \epsilon,$ however this is as far as i was able to get (i had a bit more working but none of it got me much further towards a solution).

My questions are: (1) how is $(n-N)a_n \leq a_{N+1} + ... + a_n ?$ ie i simply can't wrap my head around how this inequality holds (i understand it must have something to do with the fact that a_n is decreasing) - if someone could explain this that would be great! And (2) how would one be able to recognize that (n-N)a_n <= |S_n - S_N| < epsilon would lead to the final result? (eg clues/hints from the question?) It just doesn't come to me intuitively.

Any help, (and alternative solutions) are much appreciated

Here's the answer for (1) (i.e. why $(n-N)a_n \leq a_{N+1}+a_{N+2}+\cdots +a_n$ ).

The RHS has $n-N$ terms, and each term is greater than or equal to $a_n$ , since the given sequence is decreasing, so the RHS is greater than or equal to if each term were replaced by $a_n$ (which would make it the LHS).

shongaponga · Apr 12, 2015

InteGrand said:
Here's the answer for (1) (i.e. why $(n-N)a_n \leq a_{N+1}+a_{N+2}+\cdots +a_n$ ).

The RHS has $n-N$ terms, and each term is greater than or equal to $a_n$ , since the given sequence is decreasing, so the RHS is greater than or equal to if each term were replaced by $a_n$ (which would make it the LHS).

Ahhhhh now i see it, thanks!

Decreasing sequence of positive numbers and convergent series (1 Viewer)

shongaponga

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InteGrand

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shongaponga

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