Leibniz test $\sum\limits_{n=1}^\infty \sin\left(\pi \sqrt{n^2+\alpha}\right)$ 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?
 A: Your series is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin \left( \pi \sqrt{n^2+\alpha^2 }\right)\\
&=\sin \left( \pi n \:\sqrt{1+\frac{\alpha^2}{n^2}}\right)\\
&=\sin \left( \pi n \:\left(1+\frac{\alpha^2}{2n^2}+\mathcal{O}\left(\frac{1}{n^4}\right)\right)\right)\\
&=\sin \left( \pi n +\frac{\pi\alpha^2}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\
&=(-1)^n\sin \left(\frac{\pi\alpha^2}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\
&=\frac{\pi\alpha^2}2\frac{(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)
\end{align}
$$
giving the convergence of the initial series $\displaystyle \sum u_n$. 
A: 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.
