Using Leibniz on $\sum_{n=1}^\infty \sin(\pi \sqrt{n^2+1})$

Question

Using Leibniz on $\sum_{n=1}^\infty \sin(\pi \sqrt{n^2+1})$

So the question actually is how to rewrite $\sin(\pi\sqrt{n^2+1})$ in the form of $(-1)^n\times a_n$ so that I can apply Leibniz and decide the convergence or divergence?

I'm sorry but I'm pretty new in studying series.

Answer 1

Let $\sqrt{n^2+1} =n+d(n)$. Then

$\begin{array}\\ \sin(\pi\sqrt{n^2+1}) &=\sin(\pi(n+d(n))\\ &=\sin(\pi n)\cos(\pi d(n)) +\cos(\pi n)\sin(\pi d(n))\\ &=(-1)^n\sin(\pi d(n))\\ \end{array} $

Also

$\begin{array}\\ d(n) &=\sqrt{n^2+1}-n\\ &=(\sqrt{n^2+1}-n)\dfrac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}\\ &=\dfrac{1}{\sqrt{n^2+1}+n}\\ \end{array} $

so that $0 < d(n) \lt \dfrac1{2n}$ and $d(n)$ is decreasing so that, since $\sin(x)$ is increasing for $-\pi/2 < x < \pi/2$, $\sin(\pi d(n))$ is also decreasing.

Therefore $\sum_{n=0}^{\infty} (-1)^n \sin(\pi d(n)) $ converges by the alternating series test.

Answer 2

$$\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(\pi n\sqrt{1+\frac1{n^2}}\right)\sim\sin\left(\pi n\left(1+\frac1{2n^2}\right)\right)=(-1)^n\sin\left(\frac\pi{2n}\right)\sim(-1)^n\frac{\pi}{2n}.$$