# Using the root test when the limit does not exist

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.

• 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). – robjohn Nov 17 '18 at 14:41
• 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. – Htamstudent Nov 17 '18 at 18:31
• $\limsup$ 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. – 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.

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