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Let $\sum a_n=a$ with terms non-negative. Let $ s_n$ the n-nth partial sum. Prove $\sum na_n$ converge if $\sum (a-s_n)$ converge

El Chapo
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4 Answers4

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It is easy to prove the following fact:

If $(a_n) $ is decreasing sequence and $\sum_{j=1}^{\infty } a_j $ converges then $\lim_{j\to \infty } ja_j =0.$

Now let $r_n =\sum_{k=n+1}^{\infty } a_n $ and assume that $\sum_{i=1}^{\infty} r_n <\infty .$ Observe that $(r_n )$ is an decreasing sequence, hence by the fact $\lim_{n\to\infty} nr_n =0.$ Take any $\varepsilon >0$ and let $n_0 $ by such a big that $nr_n <\varepsilon $ and $\sum_{k=n}^{\infty} r_k <\varepsilon $ for $n\geqslant n_0 -1$ then we have $$\sum_{j=n_0}^{\infty} ja_j = n_0 r_{n_0 -1} + \sum_{k=n_0}^{\infty } r_k\leqslant \varepsilon +\varepsilon =2\varepsilon .$$ Thus the series $$\sum_{j=1}^{\infty} ja_j $$ converges.

4

This is less rigorous than the other answer but more intuitive to me (and tells us a neat thing).

$\displaystyle \sum_1^{N} (a-s_n)=Na-\sum_1^N(N+1-n)a_n$.

As $N \to \infty$, $\displaystyle Na-\sum_1^N(N+1-n)a_n \to -\sum_1^{\infty}(1-n)a_n = \sum_1^{\infty} na_n-a$.

Hence $\displaystyle \sum_1^{\infty} (a-s_n)$ converges $\displaystyle \iff \sum_1^{\infty} na_n$ converges.

The neat thing is that assuming convergence, $\displaystyle \sum_1^{\infty} (a-s_n) = \sum_1^{\infty} na_n-a$

ShakesBeer
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  • The line starting with "As $N\to\infty$" is not explained. Actually, $$\sum_1^{\infty} (a-s_n) = \sum_1^{\infty} (n-1)a_n= \left(\sum_1^{\infty} na_n\right)-a$$ (And, as always, the upvotes on mse...) – Did Jul 16 '15 at 08:16
  • @Did thanks for pointing out the minus sign, I fixed that. As for not explained, it simply depends on your personal preference. I was keeping this intuitive, and in my opinion the manipulations are fairly obvious. – ShakesBeer Jul 16 '15 at 08:40
  • Do you think the simple vanishing of $Na$ is "intuitive" and "fairly obvious"? – Did Jul 16 '15 at 08:43
  • @Did In the context of the line, yes, because you can see what cancels and what remains compared with the limit immediately to the right. At least in my opinion, comparing the two it becomes obvious that I pulled out the $\sum Na_n$ and cancelled in the limit (this being of course the least rigorous step, and perhaps even just plain wrong). – ShakesBeer Jul 16 '15 at 08:46
  • You probably mean $(n-1)a_n$, not $na_n$. Anyway, the argument relies on the fact that $$Na-\sum_1^NNa_n\to0,$$ a statement which is roughly as difficult/easy as the question itself. The answer would be better if this step was explained. Or, one can also solve the whole question saying it is "fairly obvious"... :-) – Did Jul 16 '15 at 08:52
  • @Did yes this just occured to me, and I was about to include it in an edit to my comment - but I uphold that I think it was obvious what I tried to do – ShakesBeer Jul 16 '15 at 08:52
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Simply note

$$ \sum_n (a-s_n)=\sum_n \sum_{k=n+1}^\infty a_k = \sum_{k=2}^\infty \sum_{n=1}^{k-1} a_k =\sum_{k=2}^\infty (k-1)a_k. $$

Here, we can interchange the same since all commands are nonnegative.

Using the assumption that $\sum_n a_n$ converges, this easily implies the claim (even with if and only if).

PhoemueX
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$$ \begin{align} \sum_{k=1}^nka_k &=\sum_{k=1}^n\sum_{j=1}^ka_k\\ &=\sum_{j=1}^n\sum_{k=j}^na_k\\ &=\sum_{j=1}^n(s_n-s_{j-1}) \end{align} $$ Taking limits, we get by Monotone Convergence that $$ \begin{align} \sum_{k=1}^\infty ka_k &=\sum_{j=1}^\infty(a-s_{j-1})\\ &=a+\sum_{j=1}^\infty(a-s_j)\\ \end{align} $$

robjohn
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