How prove this summation prove that:
$$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\dfrac{n}{2}$$
I think can prove  by  the  probability
my idea:
$$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\displaystyle\sum_{k=1}^{n}\dfrac{(n!)^2}{(n-k)!(n+k)!}$$
and
$$\displaystyle\sum_{k=1}^{n}\dfrac{(n!)^2}{(n-k)!(n+k)!}=(n!)^2\displaystyle\sum_{k=1}^{n}\dfrac{1}{\Gamma{(n-k+1)}\Gamma{(n+k+1)}}$$
and  use:http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Unendliche_Reihen:_Hypergeometrische_Reihen
and  my student methods:
$$\Longleftrightarrow \displaystyle\sum_{k=1}^{n}\dfrac{kC_{n}^{k}}{C_{n+k}^{k}}=\dfrac{n}{2}$$
and we notice
$$\dfrac{kC_{n}^{k}}{C_{n+k}^{k}}=\dfrac{kC_{2n}^{n+k}}{C_{2n}^{n}}$$
$$\Longleftrightarrow \displaystyle\sum_{k=0}^{n}kC_{2n}^{n+k}=\dfrac{n}{2}C_{2n}^{n}$$
However,we have 
$$\displaystyle\sum_{k=0}^{n}kC_{2n}^{n+k}=\displaystyle\sum_{k=0}^{n}C_{2n}^{k}-\displaystyle\sum_{k=0}^{n}kC_{2n}^{k}=\dfrac{n}{2}C_{2n}^{n}$$,done！
I wish to see other methods.
 A: Here is another method using the convolution of two generating functions.
Observe that when we  multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} 
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0} 
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0} 
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0} 
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the  product of  the two generating  functions is  the exponential
 generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Re-write your sum as follows:
$$\sum_{k=0}^n \frac{k\times (n!)^2}{(n+k)! (n-k)!}
= n! \sum_{k=0}^n {n\choose k} \frac{k\times k!}{(n+k)!}.$$
We proceed to evaluate the sum term.

In the present case we have
$$A(z) = \sum_{k\ge 0} \frac{k\times k!}{(n+k)!} \frac{z^k}{k!}
= \sum_{k\ge 1} \frac{k}{(n+k)!} z^k$$
and $$B(z) = \sum_{k\ge 0} \frac{z^k}{k!} = \exp(z).$$
Simplifying $A(z)$ we see that it is given by
$$z \sum_{k\ge 1} \frac{k}{(n+k)!} z^{k-1}
= z \frac{d}{dz} \sum_{k\ge 1} \frac{z^k}{(n+k)!}
= z \frac{d}{dz} \frac{1}{z^n} \sum_{k\ge 1} \frac{z^{n+k}}{(n+k)!}
= z \frac{d}{dz} 
\frac{1}{z^n} \left(\exp(z)-\sum_{q=0}^n \frac{z^q}{q!}\right)
\\= z \times
\left(-\frac{n}{z^{n+1}}
\left(\exp(z)-\sum_{q=0}^n \frac{z^q}{q!}\right)
+\frac{1}{z^n} 
\left(\exp(z)-\sum_{q=0}^{n-1} \frac{z^q}{q!}\right)\right).$$
Proceeding to extract coefficients from $A(z) B(z)$ we get
$$[z^n] A(z) B(z)
\\= - n [z^{2n}] 
\left(\exp(2z)-\exp(z)\sum_{q=0}^n \frac{z^q}{q!}\right)
+ [z^{2n-1}] 
\left(\exp(2z)-\exp(z)\sum_{q=0}^{n-1} \frac{z^q}{q!}\right)
\\ =
-n\frac{2^{2n}}{(2n)!}
+n\sum_{q=0}^n \frac{1}{q!}\frac{1}{(2n-q)!}
+ \frac{2^{2n-1}}{(2n-1)!}
-\sum_{q=0}^{n-1} \frac{1}{q!}\frac{1}{(2n-1-q)!}
\\ =
n\sum_{q=0}^n \frac{1}{q!}\frac{1}{(2n-q)!}
-\sum_{q=0}^{n-1} \frac{1}{q!}\frac{1}{(2n-1-q)!}
\\ =
\frac{n}{(2n)!} \sum_{q=0}^n {2n\choose q}
- \frac{1}{(2n-1)!} \sum_{q=0}^{n-1} {2n-1\choose q}
\\ =
\frac{n}{(2n)!} \left(2^{2n-1}+\frac{1}{2}{2n\choose n}\right)
- \frac{1}{(2n-1)!} 2^{2n-2}
\\ =
\frac{2^{2n-2}}{(2n-1)!} 
+ \frac{n}{2} \frac{1}{(2n)!} {2n\choose n}
- \frac{1}{(2n-1)!} 2^{2n-2}
= \frac{n}{2} \frac{1}{(n!)^2}.$$
The end result is that
$$n! \times n! \times [z^n] A(z) B(z)
= n! \times n! \times \frac{n}{2} \frac{1}{(n!)^2}
= \frac{n}{2}$$
as claimed. QED.
