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Show that the sum of the series $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+.....$ equal to $1+\frac{3e}{2}$.

I don't know how to find sum of this series, please tell me how to solve it?

user159480
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  • See also : https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee May 24 '17 at 05:28
  • Also: https://math.stackexchange.com/questions/1364387/finding-the-sum-of-a-series-frac11-frac122-frac1233, https://math.stackexchange.com/questions/1597328/find-the-summation-frac11-frac122-frac1233-cdots, https://math.stackexchange.com/questions/1574776/what-is-the-value-of-the-series – all found quickly using Approach0 – Martin R May 24 '17 at 05:28

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HINT:

So the $n(\ge1)$th term is $$=\dfrac{1+2+\cdots+n}{n!}=\dfrac{n(n+1)}{2(n!)}$$

Now $n(n+1)=n(n-1)+2n$

So for $n\ge2,$ $$\dfrac{n(n+1)}{n!}=\dfrac1{(n-2)!}+\dfrac2{(n-1)!}$$

Now recall $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$

lab bhattacharjee
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