I came across this result randomly and am quite sure it's right. Is there any way to prove it rigorously? The numerator always seems to be one less than the denominator. Thanks!
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This is so obviously false… – José Carlos Santos Mar 21 '19 at 08:57
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Are you sure you wrote the exact equation you wanted to write down? – Jakobian Mar 21 '19 at 08:58
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3Do you maybe mean $n = \infty$? – Dirk Mar 21 '19 at 08:58
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Indeed n = infinity – E Fresher Mar 21 '19 at 09:00
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Since $\frac{i}{(i+1)!} = \frac{(i+1)-i}{(i+1)!} = \frac{1}{i!} - \frac{1}{(i+1)!}$, your sum is a telescoping sum. – achille hui Mar 21 '19 at 09:03
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https://math.stackexchange.com/questions/845464/determine-the-value-of-the-series-sum-n-1-infty-frac1-n2-n – lab bhattacharjee Mar 21 '19 at 10:21
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$\sum\limits_{i=1}^{\infty} \frac i {(i+1!)}=\sum\limits_{i=1}^{\infty} \frac {i+1} {(i+1)!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}=\sum\limits_{i=1}^{\infty} \frac 1 {i!} -\sum\limits_{i=1}^{\infty} \frac 1 {(i+1)!}$ If just write the terms you will see that all terms cancel out except $1$.
Kavi Rama Murthy
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