real analysis - Convergence of $sumlimits_{n=1}^{infty}a_n$ implies convergence of $sumlimits_{n=1}^{infty}a_n^{sigma_n}$ where $sigma_n=frac{n}{n+1}$?

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.