Hi,
I just can't think of a way to calculate this sum:
Any help or a clue would be appriciated!
$\displaystyle \sum_{k=0}^n{n\choose k}\frac{k!}{(n+k+1)!} = \sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{k!}{(n+k+1)!} = \frac{n!}{(2n+1)!}\sum_{k=0}^n\frac{(2n+1)!}{(n-k)!(n+k+1)!}$.
Can you see why I inserted the (2n+1)! on the top and bottom of the fraction, and where to take it from there?
Thanks for your answer! I understand that now $\displaystyle \frac{(2n+1)!}{(n-k)!(n+k+1)!}$ is a binomial $\displaystyle {2n+1 \choose n-k}$ but I don't know what to do next... I've tried using the Vandermonde's identity for $\displaystyle {k \choose k}{2n+1 \choose n-k}$ but it doesn't work since k is not a constant integer...Can you tell me what should I do now?
$\displaystyle \sum_{k=0}^n{2n+1 \choose n-k}=$
$\displaystyle {2n+1 \choose n}+{2n+1 \choose n-1}+\dots+{2n+1 \choose 1}+{2n+1 \choose 0}=$
$\displaystyle {2n+1 \choose 0}+{2n+1 \choose 1}+\dots+{2n+1 \choose n-1}+{2n+1 \choose n}=$
$\displaystyle \sum_{k=0}^n{2n+1 \choose k}$
Or shift the index directly by replacing k with n-k:
$\displaystyle \begin{aligned}\displaystyle \sum_{0\le k \le n}\binom{2n+1}{n-k} & = \sum_{0\le n-k \le n}\binom{2n+1}{n-(n-k)} \\& = \sum_{-n\le -k \le 0}\binom{2n+1}{n-n+k} \\& = \sum_{0\le k \le n}\binom{2n+1}{k}. \end{aligned}$