Determine if this series : $\sum_{n=1}^\infty e^{-\sqrt{(n+1)}}$ converges or diverges.

Question

Recently i started working with infinite series and i came along with this particular series : $$\sum_{n=1}^\infty e^{-\sqrt{(n+1)}}$$ I need to prove if this infinite series converges or diverges using Direct comparison test but i do not even know where to start. Any hint would be very helpful.

Answer 1

Note that eventually

$$e^{-\sqrt{(n+1)}}=\frac1{e^{\sqrt{(n+1)}}}\le\frac1{n^2}$$

Answer 2

$e^{u}\geq u^{4}$ for large $u>0$, so $e^{-\sqrt{n+1}}\leq\dfrac{1}{(n+1)^{2}}$ and we have $\displaystyle\sum\dfrac{1}{(n+1)^{2}}<\infty$.

Answer 3

HINT: $e^{-\sqrt{n+1}}<\frac1{n^2}$ for large $n$.

Answer 4

hint: $e^{-\sqrt{n+1}} < e^{-\sqrt{n}} < \dfrac{1}{n^{3/2}}$

Answer 5

For any $u\in\mathbb{R}$ we have $e^u \geq 1+u$ (by convexity, if you like), hence we also have $$ e^u = \left(e^{u/4}\right)^4 \geq \left(1+\frac{u}{4}\right)^4 $$ or $e^{u/4}\geq \frac{e}{4}u$, implying $e^u\geq \left(\frac{e}{4}\right)^4 u^4$ and $$ \sum_{n\geq 1}e^{-\sqrt{n+1}}\leq \left(\frac{4}{e}\right)^4 \sum_{n\geq 1}\frac{1}{(n+1)^2} = \left(\frac{4}{e}\right)^4\left(\frac{\pi^2}{6}-1\right) <\pi.$$