Find convergence of $\sum_{1}^{\infty} \sin(\sqrt{n + 3} - 2\sqrt{n+2} + \sqrt{n+1})$

Question

$\sum_{1}^{\infty} \sin(\sqrt{n + 3} - 2\sqrt{n+2} + \sqrt{n+1})$

What test do I use to find out if the series is convergent or divergent? Thanks a lot!

Answer 1

$$\sqrt{n+3}-2\sqrt{n+2}+\sqrt{n+1}=$$ $$=\frac{1}{\sqrt{n+3}+\sqrt{n+2}}-\frac{1}{\sqrt{n+2}+\sqrt{n+1}}=$$ $$=-\frac{2}{(\sqrt{n+1}+\sqrt{n+2})(\sqrt{n+1}+\sqrt{n+3})(\sqrt{n+2}+\sqrt{n+3})},$$ which says that it's convergent.

Answer 2

Hint. Recall that, by Taylor expansion, $\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+o(t^2)$. Then, by letting $x=1/n$ we get $$\sqrt{n+3}-2\sqrt{n+2}+\sqrt{n+1}=\sqrt{n}\left(\sqrt{1+3x}-2\sqrt{1+2x}+\sqrt{1+x}\right)\\ =\sqrt{n}\left(1+\frac{3x}{2}-\frac{(3x)^2}{8}-2\left(1+\frac{2x}{2}-\frac{(2x)^2}{8}\right)+1+\frac{x}{2}-\frac{x^2}{8}+o(x^2)\right)\\ \sim\sqrt{n}\cdot\frac{-1}{4n^2}=- \frac{1}{4n^{3/2}}.$$ Moreover note that $\sin(t)\sim t$ as $t\to 0$. Can you take it from here?

Answer 3

$\sqrt{x}$ is a positive, increasing and concave function on $\mathbb{R}^+$. It follows that for any $n\geq 1$ the term $$ \sqrt{n+3}-2\sqrt{n+2}+\sqrt{n+1} $$ is negative, but it simple to check it is never $\leq -1$. On the other hand over the interval $[-1,0]$ the sine function is bounded between $x$ and $\frac{2}{\pi}x$, always by convexity. It follows that the given series is bounded between $K$ and $\frac{2}{\pi}K$, with $$ K = \sum_{n\geq 1}\left(\sqrt{n+3}-2\sqrt{n+2}+\sqrt{n+1}\right) \stackrel{\text{Telescopic!}}{=}\sqrt{2}-\sqrt{3}.$$ Convergence is now trivial.