The series is $\sum_{k=1}^{\infty} \frac{k}{(k+1)!}$. I can only deal with those which can be transformed into definite integral and those which have a explicit formula of summation. Any hint towards this one would be appreciated!
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1Telescoping series. – xbh Nov 14 '18 at 05:16
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3$$\frac{k+1-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$$ – Chinnapparaj R Nov 14 '18 at 05:17
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See https://math.stackexchange.com/questions/581603/evaluate-frac13-frac14-frac12-frac15-frac13-dots https://math.stackexchange.com/questions/44113/whats-the-value-of-sum-limits-k-1-infty-frack2k – lab bhattacharjee Nov 14 '18 at 05:30
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It is $$\sum_{k=1}^\infty\frac{(k+1)-1}{(k+1)!}=\sum_{k=1}^\infty\left[\frac{k+1}{(k+1)!}-\frac1{(k+1)!}\right].$$ Can you now finish off?
Angina Seng
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