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This question already has an answer here:

We have that $(b_n)$ is a sequence of decreasing, non-negative real terms. We wish to show that if $\displaystyle \sum_{i=1}^{\infty} b_n$ converges then it must be the case that $$\lim_{k \to \infty} k.b_k = 0$$

I'm stuck on this problem, I want to show this using a contradiction (assuming the limit is not zero) and showing that that contradicts the Cauchy Criterion.

Thanks.

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marked as duplicate by Antonio Vargas, Calle, TZakrevskiy, N. F. Taussig, user147263 Oct 21 '15 at 22:39

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  • $\begingroup$ It is a well known theorem.. $\endgroup$ – Empty Oct 21 '15 at 17:47
  • $\begingroup$ It isn't well known to me - the fact it is well know doesn't help me prove it unfortunately :( $\endgroup$ – Gregory Peck Oct 21 '15 at 17:51
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Hint: Cauchy Condensation test.Can you conclude using this?

Edit: $\displaystyle \sum 2^n b_{2^n} $ converges since $\displaystyle \sum b_n $ converges, and thus $ 2^n b_{2^n} \to 0. $ Now for $ 2^n < k < 2^{n+1} $,

$$ 2^n b_{2^{n+1}} \leq k b_{k} \leq 2^{n+1} b_{2^n}$$

so $n b_n \to 0.$

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Let $a_n=b_{n}-b_{n+1}\ge 0$. Then $\sum b_n<\infty$, implies that $b_n\to 0$ and hence $a_n\to 0$. Now $$ b_n=(b_n-b_{n+1})+(b_{n+1}-b_{n+2})+\cdots=\sum_{k=n}^\infty a_k, $$ and hence $$ \sum_{n=1}^\infty b_n=\sum_{k=1}^\infty ka_k. $$ But as $\sum_{k=0}^\infty ka_k<\infty$, then $\lim_{n\to\infty} \sum_{k=n}^\infty ka_k=0$. However $$ nb_n=\sum_{k=n}^\infty na_k \le \sum_{k=n}^\infty ka_k, $$ and therefore $nb_n\to 0$.

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