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Question Number 106365 by Mikael_786 last updated on 04/Aug/20

help

\underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n - 1)!}

Commented by Dwaipayan Shikari last updated on 04/Aug/20

o r s i n h (1)

Answered by Dwaipayan Shikari last updated on 04/Aug/20

Commented by Mikael_786 last updated on 04/Aug/20

thanks Sir

Commented by Ar Brandon last updated on 04/Aug/20

��wow ! What triggered you into using this identity, Shikari ? Or it's just from experience ?��

Commented by Dwaipayan Shikari last updated on 05/Aug/20

It includes factorial series .So I assumed exponential series may be useful.��

Commented by Ar Brandon last updated on 05/Aug/20

Alright, cool �� That was brilliant ��

Answered by mathmax by abdo last updated on 04/Aug/20

we have e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} and e^{- x} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n}}{n!} \Rightarrow

e^{x} - e^{- x} = \sum_{n = 0}^{\infty} \frac{1}{n!} (1 - {(- 1)}^{n}) x^{n} = \sum_{n = 0}^{\infty} \frac{2 x^{2 n + 1}}{(2 n + 1)!} \Rightarrow

\frac{e^{x} - e^{- x}}{2} = \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)!} =_{n = p - 1} \sum_{p = 1}^{\infty} \frac{x^{2 p - 1}}{(2 p - 1)!} (= sh (x))

x = 1 we get \frac{e - e^{- 1}}{2} = \sum_{p = 1}^{\infty} \frac{1}{(2 p - 1)!} \Rightarrow

\sum_{p = 1}^{\infty} \frac{1}{(2 p - 1)!} = \frac{e - \frac{1}{e}}{2} = \frac{e^{2} - 1}{2 e} (= sh (1))