1
$\begingroup$

The question is:

Show that if the partial sums $s_n$ of the series $\sum\limits_{k=1}^\infty a_k$ satisfy $\vert{s_n}\vert\leq Mn^r$ for some $r<1$, then the series $\sum\limits_{n=1}^\infty a_n/n$ converges.

My Attempt:

Let $m>n, m,n\in \mathbb{N}$ By Abel's Lemma, $$\sum\limits_{k=n+1}^m \frac{a_k}k=\frac{s_m}m-\frac{s_n}{n+1}+\sum\limits_{k=n+1}^{m-1} \left(\frac1k-\frac1{k+1}\right)s_k$$

Now I have to use the condition $\vert{s_n}\vert\leq Mn^r$ but don't know how to do so.. Could somebody help me with the solution? Or if there is a better way of doing it, could you teach me it?

$\endgroup$
  • $\begingroup$ Presumably you mean $a_k/k$ on the left side of the last equation. $\endgroup$ – Thomas Andrews Jun 16 '14 at 19:25
  • $\begingroup$ And $\frac{1}{k} - \frac{1}{k+1}$ in the sum on the right. $\endgroup$ – Daniel Fischer Jun 16 '14 at 19:27
1
$\begingroup$

Hint: $$\left|\frac{1}{k}-\frac{1}{k+1}\right|<\frac{1}{k^2}$$

$\endgroup$
0
$\begingroup$

$\left|s_{n}\right|=\left|\sum\limits_{k=1}^{n}a_{k}\right|\leq M\cdot n^{r}\quad\Rightarrow\quad \left|\frac{1}{n}\sum\limits_{k=1}^{n}a_{k}\right|\leq M\cdot n^{r-1}$, where $r-1<0$.

Hence $\left|\frac{1}{n}\sum\limits_{k=1}^{n}a_{k}\right|\to 0$, if $n\to\infty$. Now use the Cauchy criterion, or something like that.

$\endgroup$
  • 1
    $\begingroup$ it is $\sum_{n=1}^\infty \frac 1n a_n$ not $\lim_{n\to\infty} \frac1n \sum_{i=1}^n a_n$! $\endgroup$ – daw Jun 16 '14 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.