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I want to prove the following statement:

Suppose that $\displaystyle\sum_{k=1}^{\infty}a_k$ converges, where $(a_k)_{k\in\mathbb{N}}\subseteq\mathbb{R}$ is monotone. Then $\displaystyle\lim_{k\to\infty}ka_k=0$.

I believe we have several cases.

For example, if $(a_k)_{k\in\mathbb{N}}$ is monotone increasing and there exists $k$ such that $a_k>0$, then obviously $\displaystyle\sum_{k=1}^{\infty}a_k$ is not convergent.

Then, we could conclude that if some $a_k>0$ then we can suppose that $(a_k)_{k\in\mathbb{N}}$ is monotone decreasing. By the same argument, we can conclude that if some $a_k<0$ then $(a_k)_{k\in\mathbb{N}}$ must be monotone increasing.

So, I believe we only need to take care of the case where $a_k\ge 0$ for each $k\in\mathbb{N}$ and $(a_k)_{k\in\mathbb{N}}$ is monotone decreasing (the other case would be symmetric). Any hint to prove this? I have been thinking a lot ot time...

Thanks.

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marked as duplicate by Did calculus Nov 25 '14 at 22:01

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  • $\begingroup$ Right, it suffices to consider $a_k > 0$ and $(a_k)$ monotonically decreasing (non-strictly). Then suppose $\limsup ka_k > 0$ and deduce that $\sum a_k = +\infty$. $\endgroup$ – Daniel Fischer Nov 25 '14 at 21:16
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    $\begingroup$ Useful observation: $n a_{2n}\le a_{n+1}+a_{n+2}+\cdots+a_{2n}$. $\endgroup$ – David Mitra Nov 25 '14 at 21:27
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Suppose $\{a_n\}$ monotone decreasing, with $a_n \geq 0$. By the Chauchy's Criterion for convergence there exists $n_0 \in \mathbb{N}$ such that for $n+1 > n_0$ we have

$$\frac{2n \ a_{2n}}{2} = n\ a_{2n} \leq \sum_{j=n+1}^{2n} a_{2n} \leq \sum_{j=n+1}^{2n} a_{j} < \epsilon$$

Then $\lim 2n \ a_{2n} = 0 $. Now let's show that the odd part is also zero. We have that $a_{2n+1} \leq a_{2n}$ then

$$0 < (2n+1)a_{2n+1} \leq 2n\ a_{2n} + a_{2n}$$

Using the squeeze theorem we have that $\lim (2n+1)a_{2n+1} = 0$. Now as the limit of both odd and even subsequences of $\{na_n\}$ is zero then we have the result.

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  • $\begingroup$ right, I kept trying to remember this argument ... just why do you write $n+1>n_0$ instead of $n\ge n_0$ ? $\endgroup$ – Mirko Nov 25 '14 at 21:46
  • $\begingroup$ Just to take the summation from $n+1$ to $2n$. A matter of writing only. $\endgroup$ – Aaron Maroja Nov 25 '14 at 21:47
  • $\begingroup$ @AaronMaroja I don't understand why $2na_n+a_{2n}$ converges to zero. Obviously $a_{2n}$ converges to zero. But why $2na_n\to 0$? $\endgroup$ – Surtan Nov 25 '14 at 23:40
  • $\begingroup$ Well, we have just showed that it converges before. So we are just using the conclusion $\lim 2na_{2n} = 0$. $\endgroup$ – Aaron Maroja Nov 25 '14 at 23:42
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    $\begingroup$ @AaronMaroja I see what's the problem. You should write $2na_{2n}$ where you wrote $2na_n$. Thanks. $\endgroup$ – Surtan Nov 25 '14 at 23:46

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