Hi,
Prove that the attached series converges to 1
Thank's in advance
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
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
$\displaystyle 1-\frac{1}{n}\leq s_n\leq 1$