Convergence of alternating series $\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^n}{\ln^2(n) \sqrt[n]{n!}}$
I think it is clear that the series at least conditionally converges by the alternating series test.
But I'm not sure how to check if the series absolutely converge.
My idea is that since $\displaystyle\lim\limits_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e^{-1}$, the series' absolute convergence is the same as the
convergence of the series $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln^2(n)}$ .
Then, since it can be shown that $\displaystyle\int_2^\infty{\frac{1}{x\ln^2{x}}dx} < \infty$ , the original alternating series absolutely converge.
Can I actually do this kind of exchange between $n$ and $\sqrt[n]{n!}$ ?
Best Answer
Yes, your thinking is on the right way, and you can make it more rigorous by introducing a bound.
Since above limit exists, there exsits $n_1\in\mathbb{N^+}$, such that
$\frac{\sqrt[n]{n!}}{n}\le2e^{-1}$ for $\forall n\ge n_1$
Now, we have
$$\sum_{n=2}^{\infty}\left|\frac{(-1)^n}{\ln^2(n) \sqrt[n]{n!}}\right|=M+\sum_{n=n_1}^{\infty}\left|\frac{n}{n\ln^2(n) \sqrt[n]{n!}}\right|\le M+2e^{-1}\sum_{n=n_1}^{\infty}\left|\frac{1}{n\ln^2(n)}\right| $$
where
$$M=\sum_{n=2}^{n_1-1}\left|\frac{(-1)^n}{\ln^2(n) \sqrt[n]{n!}}\right|$$
By integral test,
$$\int_{n_1}^\infty \frac{1}{x\ln^2(x)} dx<\infty$$
So the original series is convergent.