TITLE: Finding whether $\sum \frac{1}{(3-(-1)^n)^n}$ converges
QUESTION [1 upvotes]: I tried to check this on WolframAlpha but it said "Standard Computation Time exceeded". I don't know whether it should converge or not.
With alternating series I usually try to find $|a_n|$. In this case I was a bit at a loss so I applied Cauchy's criteria directly to $a_n$:
$$\lim \sqrt[n]{\frac{1}{(3-(-1)^n)^n}} = \lim \frac{1}{3-(-1)^n} = 1/2 \lor 1/4$$
That's probably not the correct way to write this so feel free to correct... still either value is $<1$, so the series converges... but does it converge absolutely or conditionally? How to solve this?

REPLY [5 votes]: In that case since both $\sum \frac 1{2^n}$ and $\sum\frac 1{4^n}$ converge and the initial series is with positive terms you can not only tell the sum is convergent but also effectively calculate the actual value.
Indeed $\displaystyle \sum\limits_{n=0}^{\infty}\frac 1{3-(-1)^n}=\sum\limits_{n=0}^{\infty}\frac 1{2^{2n}}+\sum\limits_{n=0}^{\infty}\frac 1{4^{2n+1}}=\sum\limits_{n=0}^{\infty}\left(\frac 14\right)^n+\frac 14\sum\limits_{n=0}^{\infty}\left(\frac 1{16}\right)^n=\frac 43+\frac 14\times\frac {16}{15}=\frac 85$
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