Determine the convergence of the following series $\sum_{n=1}^{\infty} \sin\frac{\pi}{2\sqrt{n}}$ I want to determine the convergence of the following series
$\sum_{n=1}^{\infty} \sin\frac{\pi}{2\sqrt{n}}$ 
At first, when i saw this series i thought that it is alternating, since function $\sin x$ is sometimes above, sometimes under the $x$ axis. But then, when a took a better look at the argument i noticed that in this case, sine remains in the first quadrant since when $n=1 \Rightarrow \sin\frac{\pi}{2} = 1$ and after that, it just decreases to the point when we are looking for the sine of zero (when $n \rightarrow \infty)$, so it looks to me like it's going to converge, i just don't know how to prove that. I tried few tests, but none of them works, for example, when i try d'Alembert's test, i get $1$ which is undefined. So i am not really sure what to do with this.
 A: Always use asymptotic bounds to understand the terms.
Let "$[r]$" denote "$\{ x : |x| < |r| \}$" for any real $r$.
As $n \to \infty$:
  $\sin(\frac{π}{2\sqrt{n}}) \in \frac{π}{2\sqrt{n}} + [\frac{π}{2\sqrt{n}}^3] = \frac{π}{2\sqrt{n}} + [\frac{π^3}{8n\sqrt{n}}]$ by Taylor expansion.
Therefore $\sum_{n=1}^m \sin(\frac{π}{2\sqrt{n}}) \in \sum_{n=1}^m \frac{π}{2\sqrt{n}} + \left[ \sum_{n=1}^m \frac{π^3}{8n\sqrt{n}} \right]$ by triangle inequality.
Now it should be clear what happens, because there is a $c$ such that $\sum_{n=1}^m \frac{π^3}{8n\sqrt{n}}$ is bounded by $c$ for all $m$, and so it suffices to determine whether $\sum_{n=1}^m \frac{π}{2\sqrt{n}}$ converges as $m \to \infty$.
A: Note that $\sin \frac{\pi}{2 \sqrt{n}} \geq \frac{1}{\sqrt{n}}$ for all $n$, so the series diverges by comparison.
In general, if you have a series $\sum \sin(a_n)$ where $a_n \to 0$, you should always be able to determine convergence by comparing it to $\sum a_n$ or some constant multiple of $\sum a_n$.
A: We have that (the easiest approach I know to the following inequalities uses the fact that $\sin$ is concave in the given interval) $$\frac{2}{\pi}x\leq\sin(x)\leq x\quad\text{ for }x \in \left[0,\frac{\pi}{2}\right],$$
so 
\begin{array}{ccccccc}
\displaystyle\sum_{n=1}^{\infty} \frac{2}{\pi}\frac{\pi}{2n}
  &\leq&\displaystyle\sum_{n=1}^{\infty} \frac{2}{\pi}\frac{\pi}{2\sqrt{n}}
  &\leq&\displaystyle\sum_{n=1}^{\infty} \sin\frac{\pi}{2\sqrt{n}}
  &\leq&\displaystyle\sum_{n=1}^{\infty} \frac{\pi}{2\sqrt{n}}\\
\downarrow&&\downarrow&&&&\downarrow\\
\infty&&\infty&&&&\infty\\
\text{(harmonic series)}
\end{array}
and the sum diverges (actually, as the lower bound is $\infty$, the upper bound is not necessary).
I hope this helps $\ddot\smile$
A: As you observed this is a positive series so you can use any of the powerful tests for positive series, say limit comparison limit:
$$\lim_{n\to\infty}\frac{\sin\frac\pi{2\sqrt n}}{\frac\pi{2\sqrt n}}=1\;,\;\;\text{since}\;\;\lim_{n\to\infty}\frac\pi{2\sqrt n}=0\implies\;\text{since}\;\;\sum_{n=1}\frac\pi{2\sqrt n}$$
diverges, then also our series diverges.
