Investigate the convergence of Investigate convergence of the following series: $$\sum_{n=0}^\infty\left( \frac{2+(-1)^n}{\pi} \right)^n$$
Which convergence criterion shoul be applied?
 A: Hint:
$$\left|\left(\frac{2+(-1)^n}\pi\right)^n\right|\le\left(\frac3\pi\right)^n$$
A: As a sequence, it goes to 0.  As a series, it converges.   Proof of both is by comparison,  all the terms are nonnegative, and the top is either 1 or 3,  so go with the worst case scenario of 3, and you get $a_n \le (\frac 3 \pi )^n$, which is a geometric series with $r<1$, so it converges by the comparison test.
A: If $n$ is odd, then:
$$\left( \frac{2+(-1)^n}{\pi} \right)^n = \left( \frac{2+(-1)}{\pi} \right)^n
= \left( \frac{1}{\pi} \right)^n$$
If $n$ is even, then:
$$\left( \frac{2+(-1)^n}{\pi} \right)^n = \left( \frac{2+(+1)}{\pi} \right)^n
= \left( \frac{3}{\pi} \right)^n$$
Does that help?
Edit: apparently, you mean the series
$$\sum_{n=1}^{\infty} \left( \frac{2+(-1)^n}{\pi} \right)^n$$
See above to understand that the series is bounded above by $ \left( \frac{3}{\pi} \right)^n$ and this is a geometric series with ratio $3 / \pi < 1$, so...
A: The series is absolutely convergent by comparison with the geometric series with ratio $\frac{3}{\pi}$. 
In particular:
$$\sum_{n\geq 0}\left(\frac{2+(-1)^n}{\pi}\right)^n = \sum_{n\geq 0}\left(\frac{9}{\pi^2}\right)^n + \sum_{n\geq 0}\frac{1}{\pi^{2n+1}} = \frac{\pi^2}{\pi^2-9}+\frac{\pi}{\pi^2-1}.$$
