Answer
Convergent
Work Step by Step
Plug $e^x= a$ and $ dx=\dfrac{1}{a} da$
Now $\lim\limits_{k \to \infty} [\dfrac{2}{a}-\dfrac{2}{a+1}]_e^k=\lim\limits_{k \to \infty} [2(\dfrac{k}{k+1}-2(\dfrac{e}{e+1})]$
So, $(-2) \ln (\dfrac{e}{e+1}) \approx 0.63$
As per the Integral Test, the series is convergent