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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.

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    $\begingroup$ 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
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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.

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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.

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

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