# Comparison test for ${{{\int_{{1}}^{{\infty}}\frac{1-e^{-x}}{x}}\,{d}{x}}}$

Using the comparison test determine whether the following integral will converge.

$$\displaystyle{{{\int_{{1}}^{{\infty}}\frac{1-e^{-x}}{x}}\,{d}{x}}}$$

I've struggled with this for a good while now... Any tips on how to proceed?

• $\dfrac{1-e^{-x}}{x}\sim\dfrac{1}{x}$. So integral diverges. If you need strict comparsion, just write definition of limit. Feb 16 '20 at 11:29
• Hint: If $x\ge1$ then $1-e^{-x}\ge 1-1/e>0$. Feb 16 '20 at 11:29

$$\frac {1 -e^{-x}} x \geq \frac 1 x -\frac 1 {x^{2}} \geq \frac 1 {2x}$$ for $$x >2$$.