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.

  • 3
    $\begingroup$ If you had said that $\limsup\limits_{n\to\infty}\left|\,\frac{\cos(n)}2\,\right|\le\frac12$, then your statement would be correct, as $\limsup$ always exists (though it might be infinite). $\endgroup$ – robjohn Nov 17 '18 at 14:41
  • $\begingroup$ Thank you. I do not know what lim sup is, and I am reading about it now. I am in my second semester of calculus. $\endgroup$ – Htamstudent Nov 17 '18 at 18:31
  • $\begingroup$ $\limsup$ and $\liminf$ always exist (though each might be infinite). If they are equal, the $\lim$ exists; if they are not equal, the $\lim$ doesn't exist. $\endgroup$ – robjohn Nov 17 '18 at 18:50

We have that

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

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

$$\limsup_{n\rightarrow\infty}\sqrt[n]{\left|\left(\frac{\cos n}{2}\right)^n\right|}=L\le \frac12$$

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 $\lim_{n\to\infty}\sqrt[n]{|a_n|}=l$ 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.

  • $\begingroup$ Thank you. I just learned the extended criterion from you and robjohn today. $\endgroup$ – Htamstudent Nov 17 '18 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.