How Do You Evaluate the Sum of Reciprocal Factorial Pairs?

[anemone]

Here is this week's POTW:

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Evaluate \(\displaystyle \sum_{k=0}^{n}\dfrac{1}{(n-k)!(n+k)!}\).

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!Evaluate

 

[anemone]

Congratulations to the following members for their correct solution:):

1. lfdahl
2. kaliprasad
3. noops

Solution from lfdahl:

Spoiler

\[\sum_{j=0}^{2n}\binom{2n}{j} =\sum_{j=0}^{n-1}\binom{2n}{j}+\binom{2n}{n}+\sum_{j=n+1}^{2n}\binom{2n}{j}=2^{2n}\]

The symmetry of binomial coefficients can be expressed as: $\binom{2n}{j}=\binom{2n}{2n-j}$

- and this leads to the following identity:

\[\sum_{j=n+1}^{2n}\binom{2n}{j} = \sum_{j=0}^{n-1}\binom{2n}{j}\;\;\;\; \;\;\;\; (1).\]

So the sum can be written as:

\[\sum_{j=0}^{2n}\binom{2n}{j} = 2\sum_{j=0}^{n-1}\binom{2n}{j}+\binom{2n}{n}\]

and we get an intermediate result:

\[\sum_{j=0}^{n-1}\binom{2n}{j}=\frac{2^{2n}-\binom{2n}{n}}{2} \;\;\;\; \;\;\;\; (2).\]

Now, let´s look at the sum to be evaluated. First, make a change in the index: $j = n-k$:

\[\sum_{k=0}^{n}\frac{1}{(n-k)!(n+k)!} = \sum_{j=0}^{n}\frac{1}{j!(2n-j)!}\]

Next step is to multiply by $\frac{(2n)!}{(2n)!}$:

\[=\frac{1}{(2n)!}\sum_{j=0}^{n}\frac{(2n)!}{j!(2n-j)!} = \frac{1}{(2n)!}\sum_{j=0}^{n}\binom{2n}{j} \].

- now expand the sum to all $2n+1$ terms, and subtract the last $n$ terms:

\[=\frac{1}{(2n)!}\left ( \sum_{j=0}^{2n}\binom{2n}{j}-\sum_{j=n+1}^{2n}\binom{2n}{j} \right ) \]

- make use of $(1)$:

\[= \frac{1}{(2n)!}\left ( 2^{2n} - \sum_{j=0}^{n-1}\binom{2n}{j} \right )\]

- and of $(2)$ to get the result:

\[= \frac{1}{(2n)!}\left ( 2^{2n} - \left ( \frac{2^{2n}-\binom{2n}{n}}{2}\right ) \right ) =\frac{2^{2n-1}+\frac{1}{2}\binom{2n}{n}}{(2n)!}.\]

Alternate solution from noops:

Spoiler

First note that $\frac1{(n-k)!(n+k)!}=\frac1{(2n)!}\binom{2n}{n-k}=\frac1{(2n)!}\binom{2n}{n+k}$.
Now, we have $$\begin{align*} S&=\sum_{k=0}^n \binom{2n}{n+k}\\&=\sum_{k=0}^n \binom{2n}{n-k}\\
&= \sum_{k=0}^n \binom{2n}{k}\\&= \sum_{k=0}^{2n}\binom{2n}{k}-\sum_{k=n+1}^{2n} \binom{2n}{k} \\
&=2^{2n}-\sum_{k=1}^n \binom{2n}{k+n}\\&=2^{2n}+\binom{2n}{n}-S\\
&=2^{2n-1}+\frac12\binom{2n}{n} \end{align*}$$
So that $$ \sum_{k=0}^n \frac1{(n-k)!(n+k)!} = \frac{2^{2n-1}}{(2n)!}+\frac1{2 (n)!^{2}}.$$