TITLE: Using the root test when the limit does not exist
QUESTION [7 upvotes]: I used the root test for the series
$$
\sum_{n=1}^{\infty} \left(\frac{\cos n}{2}\right)^n.
$$
I showed that
$$
0 \le \left|\frac{\cos(n)}{2}\right| \le \frac{1}{2} \implies \lim_{n\to\infty}\left|\frac{\cos(n)}{2}\right| \le \frac{1}{2} < 1.
$$
By the root test, the series converges absolutely. My professor told me that the flaw here is that the limit above does not exist. I agree the limit does not exist because $\lvert\frac{\cos n}{2}\rvert$ oscillates between $0$ and $\frac{1}{2}$. However, I fail to see why my argument does not work here. She suggested that I use the comparison test and compare the series with $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$. By the comparison test, the original series converges absolutely. Is it a coincidence that the "pseudo" root test I used yielded the same answer as the comparison test? Can we say that if $\lvert a_n\rvert^{\frac{1}{n}}<1$, then  $\sum_{n=1}^{\infty} a_n$ converges absolutely?
I appreciate any help on this.

REPLY [9 votes]: The root test can be used without the sequence having a limit. Precisely,

if there exist $N$ and $c<1$ with $\sqrt[n]{|a_n|}\le c$ for all $n>N$, then the series $\sum_{n=0}^\infty a_n$ is absolutely convergent.

Indeed, in this case one can directly compare the series with a convergent geometric series. When $\lim_{n\to\infty}\sqrt[n]{|a_n|}=l$ exists and is $<1$, then the above criterion applies, because we can take $c=(l+1)/2$.
If you had used the “extended criterion” rather than stating that $\lim_{n\to\infty}\lvert\frac{\cos n}{2}\rvert\le \frac{1}{2}$, you would be right.