# About series $\sum_{n=1}^\infty\frac{a_n}{n^s}$

Can you see if (a), (b) and (c) ar correct, and give a hint for (d). Thabk you very much.

Let $$(a_n)$$ be a bounded sequence of real numbers.

(a) Show that the series $$\sum\limits_{n=1}^{\infty}\dfrac{a_n}{n^s}$$ is absolutely convergent if $$s>1$$.

(b) Give an example of a bounded sequence $$(a_n)$$ such that, the sequence above is divergent for all $$s\leq 1$$.

(c) Show that for $$a_n=(-1^n)$$, the series above is conditionally convergent for all $$s\in (0,1]$$.

(d) Give another example of a bounded sequence $$(a_n)$$ such that $$\lim_{n\to \infty}\sup |a_n|>0$$

and the series above is absolutely convergente for all $$s>0$$.

My solution

(a) There is $$M>0$$ such that $$|a_n|\leq M$$, for all $$n$$, so $$\left| \dfrac{a_n}{n^s}\right|\leq \dfrac{M}{n^s}$$ then, by comparison, the series at the right converges iff $$s>1$$.

(b) Let $$a_n=1$$, for $$n$$ even, and $$a_n=0$$ for $$n$$ odd

$$\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{n^{\color{red}s}}=\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{(2n)^s}=\dfrac{1}{2^s}\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^s}$$ this is divergent for $$s\leq 1$$.

(c) Let $$a_n=(-1)^n$$, then by Leibnitz' criterium $$\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{n^s}=\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^s}$$ if $$s\in(0,1]$$, then $$\dfrac{1}{n^s}$$ is decreasing and goes to zero. so it is convergent, but not absolutely convergent.

• Why you cannot just take $a_n=1$ for b) ? – Momo Jan 31 '17 at 4:28
• @momo you are right, hehe – John Jan 31 '17 at 4:36

For d) you might take take $a_n=1$, if $n=2^k$ and zero otherwise.

Then

$$\sum_{n=1}^\infty\frac{a_n}{n^s}=\sum_{k=0}^\infty\frac{1}{(2^s)^k}$$

which is a convergent geometric series. Also it is obvious that:

$$\lim_{n\to\infty}\sup_{k\ge n} |a_n|=1$$

If we can find a positive sequence $\{k_n\}$ such that

$$(1).......................\lim\limits_{n\to \infty}k_n = 0 \text{ and } \lim\limits_{n\to \infty}n^{k_n} = + \infty,$$

then (d) can be showed.

Let $k_n = \frac{1}{\ln(\ln n)}$, one can check that $\{k_n\}$ satisfies (1). Let $$M=\bigl\{m\in \mathrm{N} \mid \text{there exists } k \in\mathrm{N} \text{ such that } m=\min\{n \in \mathrm{N} \mid n^{\frac{1}{\ln(\ln n)}} > 2^k\} \bigr\},$$ where $\mathrm{N}$ is the set of natrual numbers. Define $$a_n = \begin{cases}1, \quad &n\in M, \\ 0, &otherwise, \end{cases}$$ then $a_n$ satisfies (d).

In fact, since $$\sum_n \frac{a_n}{n^{\frac{1}{\ln(\ln n)}}} \le \sum_k \frac{1}{2^k} = 1,$$ $\sum\limits_n \frac{a_n}{n^{\frac{1}{\ln(\ln n)}}}$ converges absolutely. For $s>0$ by $\lim\limits_{n\to \infty}n^{k_n} = + \infty$, we can take sufficiently large $n$ such that $\frac{1}{n^{\frac{1}{\ln(\ln n)}}} > \frac{1}{n^s}$, then (d) follows by comparison.