# Limit of a series

#### hedi

Hi,
Prove that the attached series converges to 1

#### Attachments

• 97.3 KB Views: 24

#### Plato

MHF Helper
Prove that the attached series converges to 1
There is a major mistake in your attachment. What is the $\Large k~?$

1 person

#### topsquark

Forum Staff
Hi,
Prove that the attached series converges to 1

Here's the sum in question:
$$\displaystyle \sum_{n}^{\infty} \dfrac{(-1)^k}{(k!)^2 {n \choose k}}$$

-Dan

#### hedi

the series should be

#### Attachments

• 90 KB Views: 5
1 person

#### Plato

MHF Helper
the series should be
Now is this correct: $$\displaystyle \Large\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{{{{( - 1)}^k}}}{{{{(k!)}^2}\dbinom{2n}{k}}}}~?$$

1 person

#### hedi

Sorry, it is my mistake.in the denominator there is n,not 2n

#### Plato

MHF Helper
Sorry, it is my mistake.in the denominator there is n,not 2n
Hedi. Why can't you get this right?
I have no appetite for spending time on something that you may change on the next post.

1 person

#### topsquark

Forum Staff
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 $$\displaystyle {n \choose k} = \dfrac{n!}{k! ~ (n - k)!}$$

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

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

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

What's next?

-Dan

#### Idea

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

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

1 person