Im looking to simplify $\sum_{a=1}^n \frac{a}{(a+1)!}$ into a formula (not looking for convergence!)

I've played around with the first values and the searched term seems to hold $6$ in it as well as $\frac{n(n+1)}{2}$ and very similar terms, but I fail in the end to find something.

A good first step should be $\sum_{a=1}^n \frac{a}{(a+1)!} = \frac{n(n+1)}{2} \cdot \sum_{a=1}^n \frac{1}{(a+1)!}$ and a second one, that might be more or less useful is $\frac{n(n+1)}{2} \cdot \sum_{a=1}^n \frac{1}{(a+1)\cdot a!}$

Thanks in advance, I hope the question is understandable.

  • 1
    $\begingroup$ I think you'll be hard-pressed to find a closed form of $\sum\limits_{a=1}^n \frac{1}{(a+1)!} = \sum\limits_{a=0}^{n+1} \frac{1}{a!} - 2$ since you will recognize the sum on the righthand side as a truncated expansion of $e^1$. However, for large $n$, the clipped end of the expansion is quite small, so you can approximate $\sum\limits_{a=0}^{n+1} \frac{1}{a!} -2 \approx e - 2$. However, I don't believe your first step is correct since $\sum_{a} f(a)g(a) \neq \sum_a f(a) \sum_a g(a)$. $\endgroup$
    – Tom
    Oct 22, 2016 at 19:06

1 Answer 1



$$\sum_{a=1}^n\frac a{(a+1)!}=\sum_{a=1}^n\frac {a+1-1}{(a+1)!}=\sum_{a=1}^n\left(\frac 1{a!}-\frac1{(a+1)!}\right)=1-\frac1{(n+1)!}$$


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