Thomas' Calculus 13th Edition

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