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Question Number 111067 by Algoritm last updated on 01/Sep/20

Commented by Her_Majesty last updated on 01/Sep/20

$$I{don}^{\prime }tthinkso$$$${lim}_{k\to \mathrm{\infty }}\frac{\underset{j=1}{\sum ^k}{j}^7}{k!}={lim}_{k\to \mathrm{\infty }}\frac{\frac{k^8}{8}+\frac{k^7}{2}+\frac{7k^6}{12}-\frac{7k^4}{24}+\frac{k^2}{12}}{k!}=0$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{\underset{j=1}{\sum ^k}{j}^7}{k!}=$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k^8}{8k!}+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k^7}{2k!}+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{7k^6}{12k!}-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{7k^4}{24k!}+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k^2}{12k!}=$$$$=(\frac{4140}{8}+\frac{877}{2}+\frac{7\times 203}{12}-\frac{7\times 15}{24}+\frac{2}{12})e=$$$$=\frac{25685}{24}e$$

Commented by Algoritm last updated on 02/Sep/20

$$\mathrm{thanks}$$

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