Vrftjkry

I used the root test for the series
$$
sum_{n=1}^{infty} left(frac{cos n}{2}right)^n.
$$
I showed that
$$
0 le left|frac{cos(n)}{2}right| le frac{1}{2} implies lim_{ntoinfty}left|frac{cos(n)}{2}right| le frac{1}{2} < 1.
$$
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 $lvertfrac{cos n}{2}rvert$ 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 $lvert a_nrvert^{frac{1}{n}}<1$, then $sum_{n=1}^{infty} a_n$ converges absolutely?
I appreciate any help on this.

We have that

$$ left|left(frac{cos n}{2}right)^nright|le frac1{2^n}$$

and $sum frac1{2^n}$ is a convergent geometric series, we don't need root test here.

Anyway we can also apply root test to the original series in the general form by limsup definition

$$limsup_{nrightarrowinfty}sqrt[n]{left|left(frac{cos n}{2}right)^nright|}=Lle frac12$$

and conclude that the series converges.

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 $lim_{ntoinfty}sqrt[n]{|a_n|}=l$ 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 $lim_{ntoinfty}lvertfrac{cos n}{2}rvertle frac{1}{2}$, you would be right.