Study convergence of $\int_{0}^{\infty} \frac{e^{\sqrt{x}}}{e^x + 1}$

Question

Study convergence of $$\int_{0}^{\infty} \frac{e^{\sqrt{x}}}{e^x + 1}$$

First of all, I can only use the comparison test or the limit comparison test, but I don't know to witch series compare it.

Is known that polynomials of grade $n \gt e^{x}$ and that $\sqrt{x} \lt x$ and so $e^{\sqrt{x}} \lt e^{x}$. Any hints how to proceed ? Thanks in advance.

Answer 1

Use $0\le e^{\sqrt{x}}\le e^{x/2}$ for $x\ge4$, $0\le\frac{e^{\sqrt{x}}}{e^x+1}<\frac12e^4$ for $x\in[0,\,4)$.

Answer 2

The function $\displaystyle f(x) := \frac{e^{\sqrt{x}}}{1+e^x}$ is clearly continous in the interval $[0,\infty)$.

As $x$ approaches $0$ it's perfectly integrable and as $x$ approaches $+\infty$ it decays as a negative exponential. So its integral is finite.