# What could this sum possibly converge to?

Consider

$$\sum_{i=0}^{\infty} \dfrac{n-i}{n!}$$

For $n \geq i$

Consider $n$ to be any natural number. I know for sure it's going to converge, but how do I write a formula for the sum?

Possible interpretation:

Find: $$\lim_{n\to \infty}\sum_{i=0}^{n} \dfrac{n-i}{n!}$$

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Is this supposed to be a finite sum? If so, you need a variable or a number to designate how many terms there are. If not, then review what "convergence" means. And: Why are you writing a sum in expanded form as well as using a $\sum$ symbol (without beginning/end/index too, of course)? –  anon Jul 23 '12 at 18:32
Sorry I edited my sum. –  jip Jul 23 '12 at 18:36
There is still a problem there. Don't you mean $$\sum\limits_{i = 0}^\infty {\frac{{i - n}}{{i!}}}$$ –  Pedro Tamaroff Jul 23 '12 at 18:38
Still doesn't make sense to fix a $n \geq$ i but having $i$ going from $0$ to infinity –  Jean-Sébastien Jul 23 '12 at 18:44
If you want $i\leq n$, then why not write it as $\sum_{i=0}^n ...$? –  Thomas Andrews Jul 23 '12 at 18:45

$$u_n= \sum_{i=0}^n \frac{n-i}{n!}$$ Then $$u_n = \frac{1}{n!} \sum_{i=0}^n i = \frac{1}{n!} \frac{n(n+1)}{2} = \frac{n+1}{2(n-1)!} \rightarrow 0$$
The index $n$ is not part of the sum, only $i$ varies. If you write the sum as addition of $n$ terms then every term has $1/n!$ in front of it. This is what confused everyone I believe. –  vanna Jul 23 '12 at 19:01
@jak That was why several of us thought you really meant $\frac{n-i}{i!}$, because it becomes really an "easy" problem with $\frac{n-i}{n!}$ since $\frac{1}{n!}$ is constant in the sum. –  Thomas Andrews Jul 23 '12 at 19:15