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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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    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)?2012-07-23
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    Sorry I edited my sum.2012-07-23
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    Why do you say it converges?2012-07-23
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    Because as long as $n<\infty$, the sum will just keep substracting 0, 1, 2, and consecutive natural numbers2012-07-23
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    There is still a problem there. Don't you mean $$\sum\limits_{i = 0}^\infty {\frac{{i - n}}{{i!}}} $$2012-07-23
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    Still, as I said in the commment I deleted (because it became irrelevent), the terms of the sum do not decay to zero as $n\to\infty$. What is the role of $n$?2012-07-23
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    No I actually do mean what I mean in the OP.2012-07-23
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    @jak Your sum won't stop at $n$, you'll get heaps of negative numbers, you know2012-07-23
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    Let me add a restriction then.2012-07-23
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    Still doesn't make sense to fix a $n \geq$ i but having $i$ going from $0$ to infinity2012-07-23
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    If you want $i\leq n$, then why not write it as $\sum_{i=0}^n ...$?2012-07-23
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    Or, we could stop the sum at $n$, Edit, beaten by Thomas :)2012-07-23
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    But we don't talk about finite sums converging, @jak. So your question is meaningless. Unless you are asking what is:$$\lim_{n\to\infty} \sum_{i=0}^n \frac{n-i}{n!}$$2012-07-23
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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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    How could you just factor 1/n! out like that?2012-07-23
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    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.2012-07-23
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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