How to test the convergence of the series $\sum_{n=1}^{\infty}(n!)^{\frac{1}{n}}$?
I think ratio test or comparison test is doable. But for ratio test, I cannot calculate the limit value.
Test for convergence of a series
convergence-divergencesequences-and-series
Related Solutions
The essence of my answer is the following example (please read below for a proper attribution, and a verification).
Example. Let $a_n=\frac{(−1)^{n-1}}n$ and $b^n=\frac{(−1)^n}n+\frac1{n\ln n}$.
In this case $\sum a_n$ converges and $\sum b_n$ diverges while $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=-1$.
Intuitively, allowing $c<0$ would only make sense if you allow the $a_n$ (and $b_n$) to take negative as well as positive values. Then we must deal with conditionally convergent series. These are "less stable under tweaks" (than series with all positive terms), and it turns out that we could tweak the alternating harmonic series (which as is known is convergent) into an alternating series which is divergent, yet the ratio of the common terms of the two series is a finite nonzero number (whether negative or positive is not that important). Here are more details.
The way I understand or interpret the question is the following.
We do not require that $a_n$ and $b_n$ are only non-negative terms.
We assume that $\lim_\limits{n\to \infty}\frac{a_n}{b_n}=c<0$, that is $-\infty<c<0$.
Could we conclude that either:
(i) $\sum a_n$ and $\sum b_n$ are both convergent,
or
(ii) $\sum a_n$ and $\sum b_n$ are both divergent?
The answer is NO, as explained below.
First, the essential modification of the usual limit comparison test is that we allow the $a_n$ and $b_n$ to take both positive and negative values. It is not really important that we allow $c<0$. Indeed, if $c<0$ we may replace $a_n$ with $-a_n$ and use that (obviously) the series $\sum a_n$ and the series $\sum-a_n=-\sum a_n$ are either both convergent or both divergent. (This procedure would of course replace $c$ with $-c$ .)
In one of the comments to her/his answer @user provided a
link to the following paper (preprint?)
The comparison test - Not just for nonnegative series
Michele Longo, Vincenzo Valori, October 2003.
It seems to me that @user's interpretation of the OP question was different than mine, and she/he seems to have not indicated the relevance (to my interpretation, at least ) of Example 7 in the paper referenced above.
Example 7. Let $a_n=\frac{(−1)^n}n$ and $b^n=\frac{(−1)^n}n+\frac1{n\ln n}$.
In this case $\sum a_n$ converges and $\sum b_n$ diverges while $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=1$.
The authors did not provide a proof, but I assume only
because the verification is easy.
(E.g. $\sum\frac1{n\ln n}$ is divergent by the integral test, since the improper integral $\int_3^\infty\frac{1\ dx}{x\ln x}$ is easily seen to be divergent, after making a substitution $u=\ln x$, a standard example in most calculus books.
Also, $\lim\limits_{n\to\infty}\frac{b_n}{a_n}=$ $1+\lim\limits_{n\to\infty}\frac{(−1)^n}{\ln n}=1$ . So $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\frac11=1$ .)
(Also, of course, just to note it one more time, if we let $a_n=-\frac{(−1)^n}n=\frac{(−1)^{n-1}}n$ then we would have that
$\lim\limits_{n\to\infty}\frac{a_n}{b_n}=-1$, while
$\sum a_n$ converges and $\sum b_n$ diverges).
Edit (addressing a comment by @helpme).
Intuitively that is correct, alternating series are to blame. More precisely, series that are not absolutely convergent (that is, series that are only conditionally convergent), by definition this means $\sum_na_n$ is convergent but $\sum_n|a_n|=\infty$ is not convergent. In such a series there will be infinitely many positive and infinitely many negative terms, but the signs may not necessarily alternate according to a $(-1)^n$ rule. Something like $1-\frac12-\frac13+\frac14-\frac15-\frac16...$.
But if one of the series is absolutely convergent (e.g. if $\sum_n|a_n|$ is convergent) and if $\lim_\limits{n\to\infty}\frac{a_n}{b_n}=c\in(-\infty,0)$
then we would necessarily have $\lim_\limits{n\to\infty}\frac{|a_n|}{|b_n|}=|c|\in(0,\infty)$ so $\sum_n|b_n|$ is convergent, which in turn implies that $\sum_nb_n$ is also convergent.
You can apply the comparison test to check absolute convergence:
\begin{align*} \left|\frac{\cos(n)}{\sin(2^{-n}) + n^{2}}\right| \leq \left|\frac{\cos(n)}{n^{2}}\right| \leq \frac{1}{n^{2}} \end{align*}
That is because $0 < \sin(2^{-n}) < 1$ for every $n\in\mathbb{N}$ and $|\cos(n)| \leq 1$.
Best Answer
Hint: $n! > 2^n \implies a_n = (n!)^{\frac{1}{n}}> 2 \implies a_n $ doesn't converge to $0$, hence your series not convergent. Hope it's clear now...