Convergence of $\sum\limits_{n=1}^{\infty}a_n$ implies convergence of $\sum\limits_{n=1}^{\infty}a_n^{\sigma_n}$ where $\sigma_n=\frac{n}{n+1}$?

Question

I really don't want spoilers for this problem, but I am wondering if my approach is correct. (I am of course looking for a solution based on "standard" stuff regarding series, not relying on some exotic theorem from which it may follow immediately.)

Let $ a_n > 0 $ for all $ n = 1, 2, 3, \ldots$ and suppose $ \sum\limits_{n=1}^{\infty} a_n $ converges. Let $ \sigma_n = \frac{n}{n+1} $.

To show that $ \sum\limits_{n=1}^{\infty} a_n^{\sigma_n} $ converges, first observe that

$$ \limsup\limits_{n \to \infty} ([a_n^{\sigma_n}]^{1/n}) = \limsup\limits_{n \to \infty} (a_n^{1/(n+1)}) = \limsup\limits_{n \to \infty} (a_n^{1/n}) $$

so since $ \sum a_n $ converges, $ \limsup \sqrt[n]{a_n^{\sigma_n}} \leq 1 $.

If $ \limsup (\sqrt[n]{a_n^{\sigma_n}}) < 1 $, the result follows (by the Root test). Suppose $ \limsup (\sqrt[n]{a_n^{\sigma_n}}) = 1 $ intsead. Then (this part may be totally wrong) since $ a_n > 0 $ and $ a_n \to 0 $, $ \lim\limits_{n \to \infty} \sqrt[n]{a_n^{\sigma_n}} = 1 $, which means

$$ \lim\limits_{n \to \infty} \dfrac{1}{\sqrt[n]{a_n}} = \liminf\limits_{n \to \infty} \dfrac{1}{\sqrt[n]{a_n}} = 1. $$

Hence one can find $ N \in \mathbf{N} $ such that for all $ n \geq N $, (i) $ 0 < a_n < 1 $ and (ii) $ 1 < \dfrac{1}{a_n^{1/(n+1)}} < 2 $. Then

$$ |a_n^{\sigma_n}| = \left|a_n^{1 - \tfrac{1}{n+1}}\right| = |a_n| \cdot \left|\dfrac{1}{a_n^{1/(n+1)}}\right| \leq 2|a_n| $$

so $ \sum\limits_{n=1}^{\infty} a_n^{\sigma_n} $ must converge by the Comparison test.

Answer 1

Re the "this part may be totally wrong", yes it is, if you consider $a_n=\frac{1}{e^{n}}$ where $\sum\limits_{n}a_n=\frac{e}{e-1}$. Since $$\sqrt[n]{a_n^{\sigma_n}}=\frac{1}{e^{\sigma_n}}\rightarrow\frac{1}{e}\ne 1$$

---

Incomplete work. I will use this statement where $b_n=a_n^{\sigma_n}$. Now, let's look at $$\frac{a_n}{b_n}=a_n^{1-\sigma_n}=a_n^{\frac{1}{n+1}} \tag{1}$$

If $\color{red}{\lim\limits_{n\rightarrow\infty}a_n^{\frac{1}{n+1}}=l >0}$, from the statement $\Rightarrow \sum\limits_{n}a_n^{\sigma_n}<\infty$

---

If $\color{red}{\lim\limits_{n\rightarrow\infty}a_n^{\frac{1}{n+1}}=0}$ then $$0<a_n^{\frac{1}{n+1}}<\varepsilon<1$$ from some $N$ onwards, then $$0<a_n^{\sigma_n}=a_n^{\frac{n}{n+1}}<\varepsilon^n<1$$ and $$0<\sum\limits_{n}a_n^{\sigma_n}= \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\sum\limits_{n=N+1}a_n^{\sigma_n}<\sum\limits_{n=0}^{N}a_n^{\sigma_n}+\sum\limits_{n=N+1}\varepsilon^n=\\ \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\varepsilon^{N+1}\sum\limits_{n=0}^{\infty}\varepsilon^n= \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\frac{\varepsilon^{N+1}}{1-\varepsilon}<\\ \sum\limits_{n=0}^{N}a_n^{\sigma_n}+\frac{1}{1-\varepsilon}<\infty$$

---

And the last case $\color{red}{\lim\limits_{n\rightarrow\infty}a_n^{\frac{1}{n+1}}}$ doesn't exists ... (tbc)

Answer 2

Surely, $a_n$ must be all positive. Then, we have the following inequality for all $n > 0$:

$a_n^{\sigma_n} \le \text{max} \left( \frac{1}{2^{n+1}}, a_n \right)^{\sigma_n} \le 2 \cdot \text{max}\left( \frac{1}{2^{n+1}}, a_n \right)$

Which leads directly to our the desired conclusion.