Question Number 140206 by rs4089 last updated on 05/May/21
$${Evaluate}\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}!} \\ $$$${here}\:{H}_{{n}} \:{is}\:{the}\:{nth}\:{harmonic}\:{number} \\ $$
Answered by Dwaipayan Shikari last updated on 05/May/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}!}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{{x}^{{n}} }{\mathrm{1}−{x}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{e}−\mathrm{1}}{\mathrm{1}−{x}}−\frac{{e}^{{x}} −\mathrm{1}}{\mathrm{1}−{x}}\right){dx}={e}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{e}^{{x}−\mathrm{1}} }{\mathrm{1}−{x}}{dx}\:\:\:\: \\ $$$$={e}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{e}^{−{z}} }{{z}}{dz}={e}\left(\gamma−{Ei}\left(−\mathrm{1}\right)\right) \\ $$$$ \\ $$