Convergence test for series with definite integral summand I was asked by a friend this problem: Show whether the series
$$\sum_{n = 1}^{\infty} \int_{0}^{1 / \sqrt{n}} \frac{\mathrm{e}^{x} - 1}{1 + x} \mathop{}\!\mathrm{d} x$$
converges.
Now the integral does not have a closed form, and the summand evidently vanishes as $n \to \infty$. I've tried to apply all convergence tests that I could find, but none seem to work in this case. Many of them (d'Alembert, Dirichlet, etc.) try to evaluate $\displaystyle \frac{a_{n + 1}}{a_{n}}$ and then do something about it, but in this case, when the summand is a definite integral, this fraction doesn't seem to simply things much... Nor does it seem to simply things to evaluate the square root $\sqrt{\lvert a_{n}\rvert}$ or the integral $\displaystyle \int_{1}^{\infty} a_{n} \mathop{}\!\mathrm{d} n$. The Cauchy condensation test evaluates $2^{n} a_{2^{n}}$, which also doesn't seem to lead anywhere...
I think the reason the aforementioned tests don't seem to work (based on my trial anyway...maybe they do and I'm just not seeing it) is that the summand is an integral. Hence I was wondering if there is a convergence test which works for series with definite integral summand?
Edit: As Jack pointed out below, there is no need for a test specifically for series with integral summand. (It's techniques and tricks combined with available tests)
 A: For $n >0$ , we have $$0\le x \le1$$
thus
$$e^x-1\ge x $$
and
$$\frac {1}{1+x}\geq \frac {1}{2}. $$
from this
$$\int_0^\frac {1}{\sqrt {n}}\frac {e^x-1}{1+x}dx\geq \frac {1}{2}\Bigl [\frac {x^2}{2}\Bigr]_0^\frac {1}{\sqrt {n}} $$
$$\geq \frac {1}{4n} $$
the series is Divergent.
A: A slightly less elementary, but more precise method. 
Let $a_n\stackrel{\rm def}{=}\int_{0}^{1 / \sqrt{n}} \frac{\mathrm{e}^{x} - 1}{1 + x} \mathop{}\!\mathrm{d} x$, for $n\geq 1$.
By setting $u=\sqrt{n}x$, we have
$$
na_n = n\int_{0}^{1 / \sqrt{n}} \frac{\mathrm{e}^{x} - 1}{1 + x} \mathop{}\!\mathrm{d} x
= \sqrt{n}\int_0^1 \frac{e^{\frac{u}{\sqrt{n}}}-1}{1+\frac{u}{\sqrt{n}}}\!\mathrm{d} u
$$
Defining $(f_n)_n$ on $[0,1]$ by $f_n(u) \stackrel{\rm def}{=}\sqrt{n}\frac{e^{\frac{u}{\sqrt{n}}}-1}{1+\frac{u}{\sqrt{n}}}$, it is easy to check that


*

*$f_n(u_0) \xrightarrow[n\to\infty]{} u_0$ for every fixed $u_0\in[0,1]$   (pointwise convergence)

*$\lvert f_n(u)\rvert \leq (e-1)u$ for all $u\in[0,1]$ and $n\geq 1$ (as $e^x - 1\leq (e-1)x$ for $x\in[0,1]$) (domination)
Thus, by the Dominated Convergence Theorem, we get
$$
n a_n \xrightarrow[n\to\infty]{} \int_0^1 \lim_{n\to\infty} f_n(u)\, du
= \int_0^1 u\, du = \frac{1}{2}$$
i.e. $$\boxed{a_n \operatorname*{\sim}_{n\to\infty} \frac{1}{2n}}$$
The divergence (and exact rate of divergence) of the series follow:
$$\boxed{\sum_{n=1}^N a_n \operatorname*{\sim}_{n\to\infty} \sum_{n=1}^N \frac{1}{2n} \operatorname*{\sim}_{n\to\infty} \frac{1}{2}\ln N}$$
A: Elementary inequalities do the job nicely: over the interval $[0,1]$ we have
$$\frac{e^x-1}{x+1}\geq \frac{x}{x+1} \geq x-x^2\tag{1}$$
and the series $\sum_{n\geq 1}\int_{0}^{1/\sqrt{n}}(x-x^2)\,dx$ is divergent by the p-test, so the original series is divergent as well.
