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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?

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    $\begingroup$ $\dfrac{1-e^{-x}}{x}\sim\dfrac{1}{x}$. So integral diverges. If you need strict comparsion, just write definition of limit. $\endgroup$
    – thing
    Feb 16 '20 at 11:29
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    $\begingroup$ Hint: If $x\ge1$ then $1-e^{-x}\ge 1-1/e>0$. $\endgroup$ Feb 16 '20 at 11:29
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$\frac {1 -e^{-x}} x \geq \frac 1 x -\frac 1 {x^{2}} \geq \frac 1 {2x}$ for $x >2$.

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