# What is the value of the series? [closed]

What is the value of the summation $$\dfrac{1}{1!} + \dfrac{1+2}{2!} + \dfrac{1+2+3}{3!} + \dots + {{1+2+3+\dots+i}\over{i!}} + \dots$$

The sum is till infinity.

## closed as off-topic by heropup, Simon S, user223391, user228113, BLAZEDec 14 '15 at 6:38

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## 1 Answer

Start with

$$e^x=\sum_{n=0}^\infty{x^n\over n!}$$

Derive to get

$$e^x=\sum_{n=1}^\infty{nx^{n-1}\over n!}$$

and... derive again to get

$$e^x=\sum_{n=2}^\infty{2{n(n-1)\over 2}x^{n-2}\over n!}$$

Now set $x=1$ and remember that ${n(n-1)\over 2}=1+2+\cdots +(n-1)$ to get

$$e=2\sum_{n=0}^\infty\left({1+\cdots+n\over n!}-{1\over n!}\right)$$ $$e=2\sum_{n=0}^\infty\left({1+\cdots+n\over n!}\right)-2e$$

$$\sum_{n=0}^\infty{1+\cdots + n\over n!}={3e\over 2}$$

• This is a brilliant solution. Though I still wonder: where did you find the insight to differentiate e^x twice? – zz20s Dec 14 '15 at 16:24
• Each time you have $n(n-1)\cdots (n-k)/n!$ in a series you should look into $k+1$ derivatives of $e^x$ – marwalix Dec 14 '15 at 18:09
• better yet, start with $$x e^x = \sum_{n = 1}^{\infty} \frac {x^{n + 1}} {n!}$$ and differentiate twice to get $$(x + 2) e^x = \sum_{n = 1}^{\infty} \frac {(n + 1) n x^{n - 1}} {n!}$$ and set $x = 1$ to have $$3e = 2 \sum_{n = 1}^{\infty} \sum_{k = 1}^{n} \frac {k} {n!}$$. – user98186 Dec 14 '15 at 19:41
• @Nima Bavari you're absolutely right – marwalix Dec 14 '15 at 20:27