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.