Determine the convergence or divergence of $\sum_{n=1}^\infty \frac{1}{n^{2+\sin(\frac{n\pi}{4})}}$ I have this sum and i need to study its convergence or divergence:
$$\sum_{n=1}^\infty \frac{1}{n^{2+\sin(\frac{n\pi}{4})}}$$
Of course $1 \le 2+\sin(\frac{n\pi}{4}) \le 3 $ so: $$\lim_{n\to \infty} a_n = 0$$
Anyway i can't go on from this point. I tried the comparison test and also to use the fact that $\sin(\frac{n\pi}{4})$ assumes only certain values without any success. Any help will be much appreciated. 
 A: $$
\sum_{n=1}^\infty \frac{1}{n^{2+\sin(n\pi/4)}} = \sum_{n=1\\n\not\equiv 6 (mod\ 8)}^\infty \overbrace{\frac{1}{n^{2+\sin(n\pi/4)}}}^{(A)} + \sum_{n=1\\n\equiv 6 (mod\ 8)}^\infty \overbrace{\frac{1}{n^{2+\sin(n\pi/4)}}}^{(B)}
$$
Now note that $(A) \geq \frac{1}{n^3}$ and $(B) = \frac{1}{n}$ (in the context of $n$ restricted modulo $8$). Hence:
$$
\sum_{n=1}^\infty \frac{1}{n^{2+\sin(n\pi/4)}} \geq  \sum_{n=1\\n\not\equiv 6 (mod\ 8)}^\infty \frac{1}{n^3} + \sum_{n=1\\n\equiv 6 (mod\ 8)}^\infty \frac{1}{n} = \underbrace{\sum_{n=1\\n\not\equiv 6 (mod\ 8)}^\infty \frac{1}{n^3}}_{\geq 0} + \underbrace{\sum_{n=0}^\infty \frac{1}{8n+6}}_{=\infty}
$$
A: 
The sum diverges since considering,  the sum of all integer such satisfying $n\equiv 6\mod 8$  that is n of the form, $n = 8k+6 $ then  we have, 

$$\sum_{n=1}^\infty \frac{1}{n^{2+\sin(\frac{n\pi}{4})}}\ge \sum_{n\equiv 6 \mod 8}\frac{1}{n^{2+\sin(\frac{n\pi}{4})}}\\=\sum_{n=1}^\infty \frac{1}{(4n)^{2+\sin(\frac{8n\pi}{4}+\frac{3\pi}{2})}} = \frac1{4}\sum_{n=1}^\infty \frac{1}{n} =\infty$$
Given that, $$\sin(\frac{8n\pi}{4}+\frac{3\pi}{2}) = \cos(2n\pi)\sin(\frac{3\pi}{2}) =-1$$
So the sum diverges, 
A: All the term of sereis are positive so, it suffices to under-estimate  in other to prove the divergence,  $$\sum_{n=1}^\infty \frac{1}{n^{2+\sin(\frac{n\pi}{4})}}\ge \sum_{n\equiv 2 \mod 4}\frac{1}{n^{2+\sin(\frac{n\pi}{4})}}=\sum_{n=1}^\infty \frac{1}{(4n)^{2+\sin(\frac{4n\pi}{4}+\frac{\pi}{2})}} = \sum_{n=1}^\infty \frac{1}{(4n)^{2+(-1)^{n}}} $$
Given that, $$\sin(\frac{4n\pi}{4}+\frac{\pi}{2}) = \cos(n\pi)\sin(\frac{\pi}{2}) =(-1)^n$$
Thus, $$\sum_{n=1}^\infty \frac{1}{n^{2+\sin(\frac{n\pi}{4})}}\ge  \frac1{4}\sum_{n=1}^\infty \frac{1}{(4n)^{2+(-1)^{n}}}  \ge \sum_{n\equiv 1\mod 2}^\infty \frac{1}{n^{2+(-1)^{n}}} =\frac1{4} \sum_{n=1}^\infty \frac{1}{n}=\infty$$
Which prove the divergence of the sum. 
A: The serie is divergent. Proof:
First note that the series is has positive terms. Then I can say:
$$\sum_{k = 1}^{\infty} \frac{1}{6k}  \leq \sum_{n = 1}^{\infty} \frac{1}{n^{2 + \sin{\frac{n\pi }{4}}}}$$
But $$\sum_{k = 1}^{\infty} \frac{1}{6k} = \infty$$ then the original series is divergent.
