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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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    @ThomasAndrews, yes that's I what I want to ask! Sorry for being so vague everyone2012-07-23

1 Answers 1

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$u_n= \sum_{i=0}^n \frac{n-i}{n!}$ 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$

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    @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.2012-07-23