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Question Number 176566 by mnjuly1970 last updated on 21/Sep/22

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c a l c u l a t e : Φ = \underset{n = 0}{\sum^{\infty}} \frac{1}{(2 n + 1) . e^{4 n + 2}} = ?

w h e r e^{″} e^{″} i s e u l e r n u m b e r .

≺ s o l u t i o n ≻

Φ = \underset{n = 0}{\sum^{\infty}} \frac{1}{e^{4 n + 2}} \int_{0}^{1} x^{2 n} d x

= \frac{1}{e^{2}} \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} {(\frac{x^{2}}{e^{4}})}^{n} d x

= \frac{1}{e^{2}} \int_{0}^{1} \frac{1}{1 - {(\frac{x}{e^{2}})}^{2}} d x = \frac{1}{2 e^{2}} \int_{0}^{1} \frac{1}{1 - \frac{x}{e^{2}}} + \frac{1}{1 + \frac{x}{e^{2}}} d x

= \frac{1}{2} l n (\frac{1 + \frac{1}{e^{2}}}{1 - \frac{1}{e^{2}}}) = {t a n h}^{- 1} (\frac{1}{e^{2}})

∴ Φ = {c o t h}^{- 1} (e^{2}) m . n