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Good day, solving this problem did not cause any problems, however, I don’t understand where the monotonicity of the sequence is required

If the terms of a sequence $a_n$ are monotonic, and if $\sum_1^{\infty}a_n$ converges, show that. $\sum_{n=1}^{\infty}n(a_n-a_{n+1})$converges.

Solution:

1.$\sum_{n=1}^{\infty}n(a_n-a_{n+1})=\sum_{k=1}^{\infty}\sum_{i=k}^{\infty}(a_i-a_{i+1})$

2.$\sum_{i=k}^{n}(a_i-a_{i+1})=a_k-a_{n+1}$- telescoping series

3.$\sum_{i=k}^{\infty}(a_i-a_{i+1})=a_k$ ,then $\sum_{n=1}^{\infty}n(a_n-a_{n+1})=\sum_{n=1}^{\infty}a_n$-converges.

It seems to me that the error is in point 2

Ingix
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didenko jack
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  • point 2 is OK, and it is a case of the definition of "telescoping" – coffeemath Feb 06 '24 at 06:39
  • In general $\sum_{n=1}^{\infty}n(a_n-a_{n+1})$ may not have any value assigned to it, finite or infinity. So you cannot write this kind of proof without the monotnicity assumption. – geetha290krm Feb 06 '24 at 07:18

1 Answers1

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The assumption is step 1, which is a reordering of the terms of the infinite series $\sum_{n=1}^\infty(a_n-a_{n+1})$. You can reorder the terms and get the same (finite or infinite) sum if they are all positive, which they will be if the sequence is monotonic.

However, that doesn't always work when they are not all the same sign. Indeed, if you take a sequence where the terms individually tend to zero but the sum of all positive terms and the sum of (absolute values of) all negative terms are both infinite, you can rearrange the terms to get any sum you want.

If you try the corresponding rearrangement with a large finite upper limit you get $$\sum_{n=1}^N(a_n-a_{n+1})=\sum_{k=1}^N\sum_{i=k}^N(a_i-a_{i+1})$$ Now this telescopes to $$\sum_{k=1}^N(a_k-a_{N})=-Na_N+\sum_{k=1}^Na_k.$$ The second term tends to $\sum_{k=1}^\infty a_k$, but $\lim_{N\to\infty}Na_N$ might not exist.

Especially Lime
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