I found an old solution of a problem from the American Mathematical Monthly (79–13):

Prove for m > n > 0:

∑_{i=1}^{n} i\frac{m − n + 2i + 1}{m + n + 1}\binom{m + n + 1}{n − i} = ∑_{i=1}^{n} i2^{i−1} \frac{m − n + i}{m + n − i}\binom{m + n − i}{m}.

With s = m − n, it is sufficient to prove

∑_{k=0}^{n} \frac{s + 2k + 1}{2n + s + 1}\binom{2n + s + 1}{n − k}x^{k} = ∑_{k=0}^{n} \frac{s + k}{2n + s − k} \binom{2n + s − k}{n + s}(1 + x)^{k},

since the result follows by differentiation and x = 1.

Now the lefthand side is

∑_{k=0}^{n} \frac{s + 2n − 2k + 1}{2n + s + 1}\binom{2n + s + 1}{k}x^{n−k} = ∑_{k=0}^{n} \binom{2n + s + 1}{k}x^{n−k} − 2 ∑_{k=1}^{n} \binom{2n + s}{k − 1}x^{n−k},

the righthand side is

∑_{k=0}^{n} \frac{s + n − k}{n + s + k} \binom{n + s + k}{n + s}(1 + x)^{n−k} = ∑_{k=0}^{n} \binom{n + s + k}{k}(1 + x)^{n−k} − 2 ∑_{k=1}^{n} \binom{n + s − 1 + k}{k − 1}(1 + x)^{n−k}.

It is sufficient to prove

∑_{k=0}^{n} \binom{2n + s + 1}{k}x^{n−k} = ∑_{k=0}^{n} \binom{n + s + k}{k}(1 + x)^{n−k},

∑_{k=0}^{n−1} \binom{2n + s}{k}x^{n−1−k} = ∑_{k=0}^{n−1} \binom{n + s + k}{k}(1 + x)^{n−1−k},

but these are essentially identical (replace n by n − 1 and s by s + 1 in the first one to get the second one), and for instance the first one is easily seen by [Riordan, Combinatorial Identities, p. 47, line 7], letting p = s + 1 + n.