Finding the sum of the series $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$

Question

Deteremine the sum of the series $$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$$

So I first write down the $n^{th}$ term $a_n=\frac{\frac{n(n+1)}{2}}{n!}=\frac{n+1}{2(n-1)!}$.

So from there I can write the series as $$1+\frac{3}{2}+\frac{4}{2\times 2!}+\ldots +\frac{n+1}{2(n-1)!}+\ldots $$

I am quite sure I can do some sort of term by term integration or differentiation of some standard power series and crack this. Any leads?

You don't have the correct series exactly, though the general term is okay. The correct series is:
$$\frac{2}{2\times 0!} + \frac{3}{2\times 1!} + \frac{4}{2\times 2!}+\frac{5}{2\times 3!}+\ldots +\frac{n+1}{2(n-1)!}+\ldots $$ — wythagoras, Jun 22 '15 at 08:31
Related : http://math.stackexchange.com/questions/845464/determine-the-value-of-the-following-series — lab bhattacharjee, Jun 22 '15 at 08:41
Related question: Summation of a series help: $\sum \frac{n-1}{n!}$ — Martin Sleziak, Jun 22 '15 at 08:57

score 13 · Accepted Answer · answered Jun 22 '15 at 08:29

13

HINT: Split $$\frac{n+1}{2(n-1)!}= \frac{1}{2(n-2)!}+\frac{1}{(n-1)!}$$ and use the fact that $\sum_{n=0}^{\infty} \frac{1}{n!}=e$

answered Jun 22 '15 at 08:29

Crostul

3

That does it! The sum comes out to be $\frac{3e}{2}$ – Miz Jun 22 '15 at 08:34

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