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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.

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  • $\begingroup$ Why you cannot just take $a_n=1$ for b) ? $\endgroup$ Commented Jan 31, 2017 at 4:28
  • $\begingroup$ @momo you are right, hehe $\endgroup$ Commented Jan 31, 2017 at 4:36

2 Answers 2

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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:

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

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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.

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