Hi,

Prove that the attached series converges to 1

Thank's in advance

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- Nov 6th 2018, 07:36 AMhediLimit of a series
Hi,

Prove that the attached series converges to 1

Thank's in advance - Nov 6th 2018, 09:24 AMPlatoRe: Limit of a series
- Nov 6th 2018, 09:25 AMtopsquarkRe: Limit of a series
- Nov 6th 2018, 12:26 PMhediRe: Limit of a series
the series should be

- Nov 6th 2018, 02:04 PMPlatoRe: Limit of a series
- Nov 6th 2018, 04:09 PMhediRe: Limit of a series
Sorry, it is my mistake.in the denominator there is n,not 2n

- Nov 6th 2018, 05:57 PMPlatoRe: Limit of a series
- Nov 6th 2018, 06:45 PMtopsquarkRe: Limit of a series
So is this (finally) what the question is asking you to calculate?

$\displaystyle \lim_{n \to \infty} \sum_{k = 0}^m \dfrac{(-1)^k}{(k!)^2 {n \choose k}}$

Note that $\displaystyle {n \choose k} = \dfrac{n!}{k! ~ (n - k)!}$

So your problem is

$\displaystyle \lim_{n \to \infty} \sum_{k = 0}^m \dfrac{(-1)^k}{(k!)^2 {n \choose k}} = \lim_{n \to \infty} \sum_{k = 0}^n \dfrac{(-1)^k}{(k!)^2 \dfrac{n!}{k! ~ (n - k)!}}$

$\displaystyle = \lim_{n \to \infty} \sum_{k = 0}^n \dfrac{(-1)^k k! ~ (n - k)!}{ (k!)^2 n!}$

$\displaystyle = \lim_{n \to \infty} \sum_{k = 0}^n \dfrac{(-1)^k (n - k)!}{ k! ~ n!}$

What's next?

-Dan - Nov 7th 2018, 05:14 AMIdeaRe: Limit of a series
Let

$\displaystyle s_n=\sum _{k=0}^n \frac{(-1)^k(n-k)!}{k!n!}$

expanding we get

$\displaystyle s_n=1-\frac{1}{n}+\frac{1}{2!n(n-1)}-\frac{1}{3!n(n-1)(n-2)}+\text{...}$

the terms are decreasing in absolute value so

$\displaystyle 1-\frac{1}{n}\leq s_n\leq 1$