Evaluate-n-1-H-n-n-here-H-n-is-the-nth-harmonic-number-

Question Number 140206 by rs4089 last updated on 05/May/21

E v a l u a t e \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n!}

h e r e H_{n} i s t h e n t h h a r m o n i c n u m b e r

Answered by Dwaipayan Shikari last updated on 05/May/21

\underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n!} = \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{1}{n!} (\frac{1}{1 - x} - \frac{x^{n}}{1 - x}) d x

= \int_{0}^{1} (\frac{e - 1}{1 - x} - \frac{e^{x} - 1}{1 - x}) d x = e \int_{0}^{1} \frac{1 - e^{x - 1}}{1 - x} d x

= e \int_{0}^{1} \frac{1 - e^{- z}}{z} d z = e (γ - E i (- 1))