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Question Number 159479 by quvonnn last updated on 17/Nov/21

Answered by Tokugami last updated on 17/Nov/21

$$\frac{2}{4\times 2!}+\frac{2}{5\times 3!}+\frac{2}{6\times 4!}+\frac{2}{7\times 5!}+\frac{2}{8\times 6!}+...$$$$=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{2}{(n+2)n!}$$$$=2\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1(n+1)}{(n+2)(n+1)n!}$$$$=2\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{n+1}{(n+2)!}$$$$n+2\to n$$$$=2\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{n-1}{n!}$$$$=2\underset{n=4}{\sum ^{\mathrm{\infty }}}(\frac{n}{n!}-\frac{1}{n!})$$$$=2\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{1}{(n-1)!}-2\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}$$$$\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{1}{(n-1)!}$$$$n-1\to n$$$$=2(\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}-\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{1}{n!})$$$$=2((\cancel{\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}}-\underset{n=0}{\sum ^2}\frac{1}{n!})-(\cancel{\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}}-\underset{n=0}{\sum ^3}\frac{1}{n!}))$$$$=2(-\left(\cancel{\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}}\right)+(\cancel{\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}}+\frac{1}{3!}))$$$$=\frac{2}{3!}=\frac{2}{6}$$$$=\frac{1}{3}$$

Commented by quvonnn last updated on 17/Nov/21

$$thanhyou$$

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