Prove that if a sequence converges then $\lim_{k \to \infty} kb_k = 0$

Question

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.

It isn't well known to me - the fact it is well know doesn't help me prove it unfortunately :( — Gregory Peck, Oct 21 '15 at 17:51

Arpit Kansal · Answer 1 · 2015-10-21T18:00:16.033

3

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 $n b_n \to 0.$

edited Oct 21 '15 at 18:00

answered Oct 21 '15 at 17:51

Arpit Kansal

10,268

How to prove $2^nb_{2^{n+1}} \to 0$? – Madhan Kumar Jul 25 '21 at 18:40
Dear Madhan, use necessary condition for convergence of series. Best, – Arpit Kansal Jul 26 '21 at 07:06
Yes i figured it out. Thank you. – Madhan Kumar Jul 26 '21 at 18:25

score 2 · Answer 2 · answered Oct 21 '15 at 18:16

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 $nb_n\to 0$.

Prove that if a sequence converges then $\lim_{k \to \infty} kb_k = 0$

2 Answers2