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I want to prove that:

$$\sum\limits_{i=0}^\infty \frac{(-1)^i}{i!}=\lim_{i \to \infty}\bigg(1-\frac{1}{i}\bigg)^i$$

First I need to prove series is convergent. But the partial sums of the series is not monotone sequence.Can anyone tell me how to prove that it is convergent.I tried to show it is Cauchy sequence,but I couldn't. I'd appreciate any help.

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Have you tried just applying the ratio test? – EuYu Oct 26 '12 at 18:32
@ Eu Yu.I want to solve it without using the ratio test. – p.s Oct 26 '12 at 18:35
How exactly would you like to proceed then? You should add to the question any restrictions imposed on the problem. – EuYu Oct 26 '12 at 18:38
I want to show that it is convergent without using any concepts introduced after "Convergence"topic in Analysis – p.s Oct 26 '12 at 18:46
up vote 3 down vote accepted

Use $(\rm i)$ Leibniz's Criterion, or $(\rm ii)$ a absolute convergence considerations.

$(\rm i)$ We have that $a_k=\dfrac 1 {k!}$ is monotone descreasing with $\lim\; a_k=0$, so Leibniz's criterion says $$\tag{1}\sum_{k=0}^\infty (-1)^k\frac 1 {k!}$$


$(\rm ii)$ $$\sum_{k=0}^\infty \frac 1 {k!}=e$$


$$\tag{1}\sum_{k=0}^\infty (-1)^k\frac 1 {k!}$$

converges, since it is absolutely convergent by $(1)$.

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Thank you Peter! I didnt know about the Leibniz Criterion.How do I show it converges to the same limit as $a_n=(1-1/n)^n$? Can you give me a hint,Should I show that $\sum\limits_{i=1}^n (-1)^i/i!$ is less than a_n for some n and greater than it for some m? – p.s Oct 26 '12 at 18:42
First, you should show both sequences converge to something. Then, use the binomial theorem to expand your $a_n$ and you'll see something nice going on. It will take a little work, but you should be able to show you can make the difference between the partial sums and the binomial sum as small as possible. Now I have to go, but I'll try and do it myself, too. – Pedro Tamaroff Oct 26 '12 at 18:58
I am trying to show that the sequence $s_n-a_n$ converges to zero, but no progress yet.Does anyone know how to do that? – p.s Oct 26 '12 at 19:37
@Pilot In some hours I'll give it a shot and post the results. – Pedro Tamaroff Oct 26 '12 at 20:22
@Pilot I'm back. It'd be great if you added your solution. – Pedro Tamaroff Oct 26 '12 at 22:39

Consider series of absolute values $\sum\limits_{i=0}^n \frac{1}{i!}$ and then prove that $\sum\limits_{i=0}^{\infty} \frac{(-1)^i}{i!}$ is absolutely convergent.

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