Convergence test for series with definite integral summand

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 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: $$\boxed{\sum_{n=1}^N a_n \operatorname*{\sim}_{n\to\infty} \sum_{n=1}^N \frac{1}{2n} \operatorname*{\sim}_{n\to\infty} \frac{1}{2}\ln N}$$

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.