Evaluate the Binomial Sum

Question

Evaluate the following sum involving binomial coefficients, $$ \sum_{k=0}^n \binom{n}{k}\frac{k!}{(n+1+k)!} $$ when $n$ is a nonnegative integer.

Answer 1

$$ \eqalign{ & \sum\limits_{k = 0}^n {\left( \matrix{ n \cr k \cr} \right)} {{k!} \over {\left( {n + 1 + k} \right)!}} = \sum\limits_{k = 0}^n {{{n!} \over {\left( {n - k} \right)!\left( {n + 1 + k} \right)!}}} = \cr & = {{n!} \over {\left( {2n + 1} \right)!}}\sum\limits_{k = 0}^n {\left( \matrix{ 2n + 1 \cr n - k \cr} \right)} = {{n!} \over {\left( {2n + 1} \right)!}}\sum\limits_{k = 0}^n {\left( \matrix{ 2n + 1 \cr k \cr} \right)} = \cr & = {{n!} \over {\left( {2n + 1} \right)!}}{1 \over 2}\left( {\sum\limits_{k = 0}^{2n + 1} {\left( \matrix{ 2n + 1 \cr k \cr} \right)} } \right) = {{n!} \over {\left( {2n + 1} \right)!}}2^{\,2n} \cr} $$

Answer 2

Hint. Rewrite the term in the following way $$\binom{n}{k}\frac{k!}{(n+1+k)!}=\frac{n!}{k!(n-k)!}\cdot \frac{k!}{(n+1+k)!}=\frac{n!}{(2n+1)!}\cdot \binom{2n+1}{n-k},$$ then use the Binomial Theorem. Can you take it from here?

P.S. Take a look to Relation between sum of combinations

Answer 3

If you write a few partial sums, you can see a pattern (for a product): $$\begin{align}\sum_{k=0}^0 \binom{0}{k}\frac{k!}{(0+1+k)!}=&1\cdot \frac{1}{1}=1\\ \sum_{k=0}^1 \binom{1}{k}\frac{k!}{(1+1+k)!}=&\binom{1}{0}\frac{0!}{(1+1+0)!}+\binom{1}{1}\frac{1!}{(1+1+1)!}=\\ =&\frac 1{2!}+\frac 1{3!}=\frac12+\frac16=1\cdot \frac23;\\ \sum_{k=0}^2 \binom{2}{k}\frac{k!}{(2+1+k)!}=&\binom{2}{0}\frac{0!}{(2+1+0)!}+\binom{2}{1}\frac{1!}{(2+1+1)!}+\binom{2}{2}\frac{2!}{(2+1+2)!}=\\ =&\frac 1{3!}+\frac {2\cdot 1!}{4!}+ \frac{2!}{5!}=\frac 4{15}=1\cdot \frac23\cdot \frac25;\\ \sum_{k=0}^3 \binom{3}{k}\frac{k!}{(3+1+k)!}=&\cdots=\\ =&\frac {1}{4!}+\frac {3\cdot 1!}{5!}+\frac {3\cdot 2!}{6!}+\frac {3!}{7!}=\frac 8{105}=1\cdot \frac23\cdot \frac25 \cdot \frac 27;\end{align}$$ Hence: $$\sum_{k=0}^n \binom{n}{k}\frac{k!}{(n+1+k)!}=\frac12 \prod_{k=0}^n \frac2{(2k+1)}.$$ Note: Although it is not a closed form, but it is a more convenient form.