This question seems similar:
Find a closed form for the expression
for .
.
But I am unable to form the required integral in my problem.
Prove that:
(n!m!)/(m + n)! = (nC1)/(m + 1) - 2.(nC2)/(m + 2) + 3.(nC3)/(m + 3) - ... + (-1)^(n+1)n.(nCn)/(m + n)
NOTE: nC1 means n choose 1. I struggled with latex so I had to eventually use this notation.
Yes you are right this is the same kind of problem
-1)^{k-1}}{m+k} = \sum_{k=1}^{n} \binom{n}{k} \int_{0}^{1} k\-1)^{k-1} x^{m+k-1} dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \sum_{k=1}^{n} \binom{n}{k} \int_{0}^{1} k\-1)^{k-1} x^{m+k-1} dx" />
-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-1)^{k-1} x^{k-1} dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-1)^{k-1} x^{k-1} dx" />
-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} dx" />
Now
By differentiation with respect to x
-x)^{k-1} = n\1-x)^{n-1}" alt="\sum_{k=1}^{n} \binom{n}{k} k\-x)^{k-1} = n\1-x)^{n-1}" />
Therefore
-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \:n\1-x)^{n-1}dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = \int_{0}^{1} x^m \:n\1-x)^{n-1}dx" />
-1)^{k-1}}{m+k} = n\:\int_{0}^{1} x^m \1-x)^{n-1}dx" alt="\sum_{k=1}^{n} \binom{n}{k} \:\frac{k\-1)^{k-1}}{m+k} = n\:\int_{0}^{1} x^m \1-x)^{n-1}dx" />
Now you just need to show that
1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" alt="n\:\int_{0}^{1} x^m \1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" />