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Let $(a_k)$ be a sequence decreasing to $0$, for which $$ b_k:=\left(\sum_{l=1}^ka_l\right)-ka_{k+1}=\sum_{l=1}^k(a_l-a_{k+1})$$ is bounded. Does $\sum_{k\geq 0}a_k$ necessarily converge?

Some immediate observations:

  • $(b_k)$ is increasing, hence converges.
  • By considering $b_k-b_{k-1}$, one obtains that $k(a_k-a_{k+1})$ tends towards $0$.
  • For $\sum a_k$ to converge, it suffices to prove that $ka_k$ converges.
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We know that $(b_k)_{k\in\mathbb N}$ is convergent. Let $b$ be its limit.

Let $p$ be a positive integer.

$\forall k \geqslant p \quad , \quad b_k = \displaystyle \sum_{\ell = 1}^k (a_{\ell}-a_{k+1})\geqslant \sum_{\ell =1}^p (a_{\ell}-a_{k+1}) $

Then $\quad \lim b_k \displaystyle \geqslant \lim_{k\rightarrow +\infty} \sum_{\ell=1}^p (a_{\ell}-a_{k+1})$

So $\quad b \geqslant \displaystyle\sum_{\ell = 1}^p a_{\ell} $

And we can conclude that $\sum a_k$ is convergent.

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  • $\begingroup$ Very nice. Thanks a lot! $\endgroup$ – Zuy Mar 14 at 5:53

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