Limit of a series

Oct 2012
257
19
israel
Hi,
Prove that the attached series converges to 1

Thank's in advance
 

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topsquark

Forum Staff
Jan 2006
11,565
3,453
Wellsville, NY
Hi,
Prove that the attached series converges to 1

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

-Dan
 

Plato

MHF Helper
Aug 2006
22,469
8,640
the series should be
Now is this correct: \(\displaystyle \Large\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{{{{( - 1)}^k}}}{{{{(k!)}^2}\dbinom{2n}{k}}}}~? \)
 
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Oct 2012
257
19
israel
Sorry, it is my mistake.in the denominator there is n,not 2n
 

Plato

MHF Helper
Aug 2006
22,469
8,640
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.
 
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topsquark

Forum Staff
Jan 2006
11,565
3,453
Wellsville, NY
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
 
Jun 2013
1,110
590
Lebanon
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

\(\displaystyle 1-\frac{1}{n}\leq s_n\leq 1\)
 
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