does $\sum_{n=1}^\infty \int_{n}^{n+1} \frac{1}{e^\sqrt x } dx$ converges? Be $$\sum_{n=1}^\infty \int_{n}^{n+1} \frac{1}{e^\sqrt x } dx$$ This is convergent or not?
So applying inequalities we have:
$$n\le x\le n+1$$
$$\sqrt n\le \sqrt x\le \sqrt {n+1}$$
$$e^\sqrt n\le e^\sqrt x\le e^\sqrt {n+1}$$
$$\frac{1}{e^\sqrt n}\ge  \frac{1}{e^\sqrt x} \ge  \frac{1}{e^\sqrt {n+1}}$$
Rearranging
$$\frac{1}{e^\sqrt {n+1}}\le  \frac{1}{e^\sqrt x} \le \frac{1}{e^\sqrt n} $$
Applying integrals
$$\int_{n}^{n+1}\frac{1}{e^\sqrt {n+1}}\le \int_{n}^{n+1} \frac{1}{e^\sqrt x} \le \int_{n}^{n+1}\frac{1}{e^\sqrt n} $$
but I get stuck here, I know that I have to apply $\sum_{n=1}^\infty$ to both sides of the inequality and conclude by a comparison test but I do not how to proceed. So any help will be appreciated. 
 A: hint
We know that
$$\lim_{n\to +\infty}n^4e^{-n}=0$$
thus
$$\lim_{n\to+\infty}n^2e^{-\sqrt {n}}=0$$
and for large $n $
$$n^2e^{-\sqrt {n}}<1$$
or
$$e^{-\sqrt {n}}<\frac {1}{n^2} $$
Can you conclude.
A: One thing we could do is actually calculate the sum. First let's look at the integral
$$\int_n^{n+1}e^{-\sqrt{x}}dx.$$
To do so, set $u= \sqrt{x}$, so $du = \frac{1}{2u}dx$. The integral then becomes
$$\int_{\sqrt{n}}^{\sqrt{n+1}} 2ue^{-u} du.$$ Integrating by parts yields that this is equal to 
$$2\left(-ue^{-u}\biggm|_{\sqrt{n}}^{\sqrt{n+1}} + \int_{\sqrt{n}}^{\sqrt{n+1}} e^{-u}du\right)$$
$$= 2 \left(\sqrt{n}e^{-\sqrt{n}} -\sqrt{n+1}e^{-\sqrt{n+1}} + e^{-\sqrt{n}} -
 e^{-\sqrt{n+1}}\right)$$
Looking at partial sums we see 
$$\sum_{n=1}^N 2 \left(\sqrt{n}e^{-\sqrt{n}} -\sqrt{n+1}e^{-\sqrt{n+1}} + e^{-\sqrt{n}} -
 e^{-\sqrt{n+1}}\right)$$
$$= 2\left(e^{-1} -\sqrt{N+1}e^{-\sqrt{N+1}}+e^{-1}-e^{-\sqrt{N+1}}\right)\to \frac4e$$
as $N \to \infty$.
The partial sum follows by telescoping. 
A: $f(x)=e^{-\sqrt{x}}$ is continuous and decreasing (towards zero) on $\mathbb{R}^+$, hence
$$ 0\leq f(n+1)\leq \int_{n}^{n+1} f(x)\,dx \leq f(n) $$
and since both $\sum_{n\geq 1}e^{-\sqrt{n}}$ and $\sum_{n\geq 1}e^{-\sqrt{n+1}}$ are convergent (by asymptotic comparison with a geometric series, for instance), the same holds for the original series, by squeezing.
A: For $\quad f(x)=-2e^{-\sqrt{x}}(\sqrt{x}+1)\quad$ then $\quad f'(x)=e^{-\sqrt{x}}$
So $\displaystyle u_n=\int_{n}^{n+1}\frac{dx}{e^{\sqrt{x}}}=f(n+1)-f(n)$
Note that as an integral on $[n,n+1]$ of a positive function then $u_n>0$.
By rough majoration we have $0<u_n\le 2\times |2e^{-\sqrt{n}}(2\sqrt{n+1})|<8ne^{-\sqrt{n}}<\frac 1{n^2}$ for $n$ large enough since exponential drives any power of $n$ to $0$.
The series thus converges absolutely and we can sum it by telescoping process.
$\displaystyle \sum\limits_{n=1}^{\infty} u_n=\lim\limits_{n\to\infty}f(n)-f(1)=0-f(1)=\frac 4e$
