Limit of a series

• Nov 6th 2018, 07:36 AM
hedi
Limit of a series
Hi,
Prove that the attached series converges to 1

• Nov 6th 2018, 09:24 AM
Plato
Re: Limit of a series
Quote:

Prove that the attached series converges to 1

There is a major mistake in your attachment. What is the $\Large k~?$
• Nov 6th 2018, 09:25 AM
topsquark
Re: Limit of a series
Quote:

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
• Nov 6th 2018, 12:26 PM
hedi
Re: Limit of a series
the series should be
• Nov 6th 2018, 02:04 PM
Plato
Re: Limit of a series
Now is this correct: $\displaystyle \Large\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{{{{( - 1)}^k}}}{{{{(k!)}^2}\dbinom{2n}{k}}}}~?$
• Nov 6th 2018, 04:09 PM
hedi
Re: Limit of a series
Sorry, it is my mistake.in the denominator there is n,not 2n
• Nov 6th 2018, 05:57 PM
Plato
Re: Limit of a series
Quote:

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.
Someone may help you but it will not be me. Sorry.
• Nov 6th 2018, 06:45 PM
topsquark
Re: 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 AM
Idea
Re: 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

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