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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.



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

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 $n b_n \to 0.$


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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