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The sum of the series : $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+...$ equals

A) e

B) $\frac{e}{2}$

c) $\frac{3}{2}e$

D) $1+\frac{e}{2}$

I need a hint to deal with this sum, I understood, since the sum of the first n natural number is $\frac{n(n+1)}{2}$, so the $n^{th}$ term of the series is $\frac{\frac{n(n+1)}{2}}{n!}=\frac{n(n+1)}{2n!}$, how should I use this to get the sum...

Thanks in advance!!

math student
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    It is greater than $e$. – David Mitra Nov 03 '22 at 11:00
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    https://math.stackexchange.com/q/1334661/42969 – Martin R Nov 03 '22 at 11:01
  • Your sum = $\sum_{k=1}^\infty \frac{k(k+1)}{2k!}$... – Surb Nov 03 '22 at 11:02
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    Hint: Write $$\frac{n(n+1)}{n!}=\frac{n(n-1)}{n!}+2\frac n{n!},$$ and play around with the series $e^x=1+\sum_{n=1}^\infty \frac{x^n}{n!}$ (taking derivatives). – Feng Nov 03 '22 at 11:02
  • A word of warning, writing $2n!$ could be confusing since some people might incorrectly read that as $(2n)!$ as opposed to $2(n!)$. Better probably to use parentheses or a multiplication symbol between them to emphasize this. – JMoravitz Nov 03 '22 at 13:48

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