Check if it absolutely converges or not I'm trying to determine if $\sum \limits_{n=1}^{\infty} \sin(n\pi + \frac{1}{2n})$ absolutely converges or not.
Help me check it. I don't know how to do it. Advance thanks. :)
 A: Observe
\begin{align}
\sin\left(n\pi +\frac{1}{2n}\right) = (-1)^n\sin\frac{1}{2n}
\end{align}
then apply alternating series test. 
It should be noted for large enough $n$ the terms $\sin\frac{1}{2n}$ decreases monotonically to zero. 
Hence the series converges conditionally.
Edit: Let us consider the case of absolute convergence. Observe
\begin{align}
\sin x = x-\frac{x^3}{3!} +\mathcal{O}(x^5)
\end{align}
which means
\begin{align}
\sin x \geq x-\frac{x^3}{3!}
\end{align}
for a neigbhorhood close to $0$. Hence it follows
\begin{align}
\sum^\infty_{n=N} \sin \frac{1}{2n} \geq \sum^\infty_{n=N} \frac{1}{2n} -\frac{1}{3!}\sum^\infty_{n=N} \frac{1}{8 n^3} 
\end{align}
which shows that the series diverges absolutely because $\sum^\infty_{n=N} \frac{1}{n} = \infty$. 
A: For each $n>0$
$\sin(n\pi+\frac{1}{2n})=(-1)^n\sin(\frac{1}{2n})$
and
$$|\sin(n\pi+\frac{1}{2n})|=\sin(\frac{1}{2n}).$$
but
$$\sin(\frac{1}{2n})=\frac{1}{2n}(1+(2n\sin(\frac{1}{2n})-1))$$
$$=\frac{1}{2n}(1+\epsilon(n))$$ with
$$\lim_{n\to+\infty}\epsilon(n)=0.$$
thus for large enough $n$,
$$\frac{1}{2}\frac{1}{2n}<\sin(\frac{1}{2n})$$
and by comparison with harmonic series, $$\sum \sin(\frac{1}{2n})$$ is divergent.
A: Another approach, perhaps simpler but I think that shorter anyway: the limit comparison theorem with $\;b_n=\frac1{2n}\;$ , then
$$\frac{a_n}{b_n}=\frac{\sin\frac1{2n}}{\frac1{2n}}\xrightarrow[n\to\infty]{}1$$
and thus the series $\;\sum\sin\frac1{2n}\;,\;\;\sum\frac1{2n}\;$ converge or diverge together...but the latter is just a constant multiple of the harmonic one.
A: Notice that $|\sin(n\pi +{1\over2n})| = \sin({1\over2n}) \sim {1\over2n}$ and therefore $\sum \sin(n\pi +{1\over2n})$ is absolutely divergent (because $\sum{1\over n}$ is)
