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Question Number 154824 by talminator2856791 last updated on 21/Sep/21

$$$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{4^{-k}\mathrm{\Gamma }\left(k\right)}{k!}$$$$$$

Answered by Ar Brandon last updated on 21/Sep/21

$$S=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{4^{-k}\mathrm{\Gamma }\left(k\right)}{k!}=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{4^{-k}(k-1)!}{k!}$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{4}\right)}^k\frac{1}{k}\to \mathrm{divergent}\mathrm{as}{\left(\frac{1}{4}\right)}^k\frac{1}{k}\to +\mathrm{\infty }\mathrm{for}k=0$$

Commented by talminator2856791 last updated on 22/Sep/21

$$\mathrm{wrong}.$$

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