TITLE: Leibniz test $\sum\limits_{n=1}^\infty \sin\left(\pi \sqrt{n^2+\alpha}\right)$
QUESTION [6 upvotes]: I am given a series:
$$\sum\limits_{n=1}^\infty \sin\left(\pi \sqrt{n^2+\alpha}\right)$$
And in the description of the problem it is said that I must show by Leibniz test that it is convergent. How can I even get the $(-1)^n$ symbol out of the starting series?

REPLY [6 votes]: Write $$\begin{align}
\sqrt{n^2+\alpha}&=n + \left(\sqrt{n^2+\alpha}-n\right)\\
&=n+\frac{\alpha}{\sqrt{n^2+\alpha}+n}
\end{align}
$$
Now $\sin (n\pi + x)=(-1)^n\sin(x)$. Letting $$x_n=\frac{\alpha}{\sqrt{n^2+\alpha}+n}$$ we see that it is positive and decreases to zero, and thus that $\sin(x_n\pi)$ is positive and decreases to zero.
You might have to slightly alter the argument if $\alpha$ can be negative, but deal with that as a special case.