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Question: Suppose $a_k>0$ for $k\in\mathbb{N}$ and the series $\sum_{k=1}^{\infty} a_k$ converges. Prove that there exists a sequence $(\gamma_k)_{k=1}^{\infty}$ with $\gamma_k>0$ and $\gamma_k\to \infty$ when $n\to \infty$ such that $\sum_{k=1}^{\infty} \gamma_ka_k$ converges as well.

What did I do: I know that $\sqrt[n]{a_n}\to L$ when $n\to \infty$ for some $0\leq L\leq 1$ by the root test. If $0<L<1$, then we can choose $\gamma_n=\left(\frac{L+1}{2L}\right)^n$. By that, $\frac{L+1}{2L}>1$ and $\gamma_n\to \infty$ will be satisfied. Also, we will obtain that $\sqrt[n]{\gamma_na_n}\to \frac{L+1}{2}$ which is less than $1$. Therefore, everything will be fine. Also, if $L=0$, we can simply choose $\gamma_n=2^n$. However, I do not know what to do when $L=1$ because everything is in blur at this point. How can I continue?

lecdabster
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  • If $a_n > 0$ and $\sum a_n$ converges, then $\limsup_{n\to \infty} a_n ^{1/n} < 1$, but does that mean the limit of $a_n^{1/n}$ always exists? – RRL Sep 29 '22 at 23:18
  • Root test – RRL Sep 29 '22 at 23:23
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    @RRL yes, you are right. As a counterexample to my work, we can define $a_n=3^{-n}$ when $n$ is odd and $a_n=2^{-n}$ when $n$ is even. I always thought in this question as if $a_n$ was a continuous function like $\frac{1}{n^2}$. – lecdabster Sep 30 '22 at 09:43

3 Answers3

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Let $T_n = \sum_{k=n}^\infty a_k$. Since $\sum a_k$ converges we have $\lim_{n\to \infty} T_n = 0$. Since $a_k >0$ , the sequence $(T_n)$ is decreasing.

Taking $\gamma_n = \frac{1}{\sqrt{T_n}}$, we have $\gamma_n \to \infty$ as $n \to \infty$ and

$$\gamma_na_n = \frac{a_n}{\sqrt{T_n}} = \frac{T_n - T_{n+1}}{\sqrt{T_n}}= \frac{(\sqrt{T_n}-\sqrt{T_{n+1}})(\sqrt{T_n}+\sqrt{T_{n+1}})}{\sqrt{T_n}}\leqslant 2(\sqrt{T_n}-\sqrt{T_{n+1}})$$

Thus,

$$\tag{*}\sum_{n=N+1}^M\gamma_na_n \leqslant 2\sum_{n=N+1}^M(\sqrt{T_n}-\sqrt{T_{n+1}})=2(\sqrt{T_{N+1}}- \sqrt{T_{M+1}})$$

Since the sequence $(\sqrt{T_n})$ is a Cauchy sequence, (*) implies that the sequence of partial sums for $\sum \gamma_na_n$ is a Cauchy sequence and, therefore, the series $\sum \gamma_na_n$ converges.

RRL
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You know that the sequence $s_n \colon =\sum_{k=1}^n a_k$ is convergent, and so (easy) it is a Cauchy sequence. Therefore, we can find an sequence $N_n$ such that $|s_m - s_n| < \frac{1}{n^3}$ for $m$, $n\ge N_n$. In particular

$$s_{N_{n+1}} - s_{N_n} < \frac{1}{n^3}$$

Now you can multiply the terms of the series from $N_n$ to $N_{n+1}$ by $n$ and still get $$s'_{N_{n+1}} - s'_{N_n} < \frac{1}{n^2}$$

orangeskid
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So we have a sequence of positive reals $a_n$, with $\sum_{n=1}^{\infty} a_n=S\in \mathbb{R}$.

For any non-decreasing sequence of (strictly) positive integers $\gamma_n$, we have: $$ \sum_{n=1}^{\infty} \gamma_na_n=\sum_{k=1}^\infty\sum_{\,\,\,\,n|\,\gamma_n\geq k} a_n, $$ by which I mean either both sides converge and are equal, or both sides diverge. To see this, just note that we get an $a_n$ term on the right handside for each integer $k=1,2,\dots,\gamma_n$.

Thus taking $\gamma_n={\rm max}\left(1, \lfloor -\log(\sum_{r=n}^\infty a_n)\rfloor\right)$, we have that $\gamma_n\to\infty$ as $n\to \infty$.

Note that when $\gamma_n\geq 2$ we have $$\sum_{r=n}^\infty a_n\leq e^{-\gamma_n}.$$

Thus: \begin{eqnarray} \sum_{n=1}^{\infty} \gamma_na_n=\sum_{k=1}^\infty\sum_{\,\,\,\,n|\,\gamma_n\geq k} a_n&=&\sum_{n=1}^{\infty} a_n + \sum_{k=2}^\infty\quad\sum_{r={\rm min}\{n|\lambda_n\geq k\}}^\infty a_r \\ &\leq& \sum_{n=1}^{\infty} a_n + \sum_{k=2}^\infty e^{-k}=S+\frac1{e^2-e}. \end{eqnarray}

tkf
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