Evaluate:$\sum_{s=0}^{N}(-1)^s{2s \choose s}{N+s \choose N-s}\frac{s^2}{(s+1)^2}$

Question

We seek to evaluate $(1)$: $$S(N)=\sum_{s=0}^{N}(-1)^s{2s \choose s}{N+s \choose N-s}\frac{s^2}{(s+1)^2}\tag1$$

some simplification

$$S(N)=\sum_{s=0}^{N}(-1)^s\frac{(N+s)!}{(N-s)!(s-1)!^2}\frac{1}{(s+1)^2}$$

Answer 1

We seek to evaluate

$$S(N) = \sum_{q=0}^N (-1)^q {2q\choose q} {N+q\choose N-q} \frac{q^2}{(q+1)^2}$$

or alternatively

$$S(N) = \sum_{q=0}^N (-1)^q \frac{(N+q)!}{(N-q)!(q-1)!^2} \frac{1}{(q+1)^2}.$$

This is

$$S(N) = \sum_{q=0}^N q^2 (-1)^q \frac{(N+q)!}{(N-q)!(q+1)!^2} \\ = \sum_{q=0}^N q^2 (-1)^q {N+1\choose q+1} \frac{(N+q)!}{(N+1)! (q+1)!} \\ = \frac{1}{N(N+1)} \sum_{q=0}^N q^2 (-1)^q {N+1\choose q+1} \frac{(N+q)!}{(N-1)! (q+1)!} \\ = \frac{1}{N(N+1)} \sum_{q=0}^N q^2 (-1)^q {N+1\choose q+1} {N+q\choose q+1}.$$

We continue with

$$\frac{1}{N(N+1)} \sum_{q=0}^N q^2 (-1)^q {N+1\choose N-q} {N+q\choose q+1} \\ = \frac{1}{N(N+1)} [z^N] (1+z)^{N+1} \sum_{q=0}^N q^2 (-1)^q z^q {N+q\choose q+1}.$$

Here the coefficient extractor enforces the upper limit of the sum:

$$\frac{1}{N(N+1)} [z^N] (1+z)^{N+1} \sum_{q\ge 0} q^2 (-1)^q z^q {N+q\choose N-1} \\ = \frac{1}{N(N+1)} [z^N] (1+z)^{N+1} [w^{N-1}] (1+w)^N \sum_{q\ge 0} q^2 (-1)^q z^q (1+w)^q \\ = \frac{1}{N(N+1)} [z^N] (1+z)^{N+1} [w^{N-1}] (1+w)^N \frac{-z(1+w)(1-z(1+w))}{(1+z(1+w))^3} \\ = -\frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N+1} [w^{N-1}] (1+w)^{N+1} \frac{1-z(1+w)}{(1+z(1+w))^3}.$$

We have two pieces here, the first one is

$$-\frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N+1} [w^{N-1}] (1+w)^{N+1} \frac{1}{(1+z(1+w))^3} \\ = -\frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N-2} [w^{N-1}] (1+w)^{N+1} \frac{1}{(1+zw/(1+z))^3}.$$

The inner term is

$$\sum_{q=0}^{N-1} {N+1\choose N-1-q} (-1)^q {q+2\choose 2} \frac{z^q}{(1+z)^q}.$$

Now

$${N+1\choose N-1-q} {q+2\choose 2} = \frac{(N+1)!}{(N-1-q)! \times q! \times 2!} = {N+1\choose 2} {N-1\choose q}$$

and we find for the inner term

$${N+1\choose 2} \sum_{q=0}^{N-1} {N-1\choose q} (-1)^q \frac{z^q}{(1+z)^q} = {N+1\choose 2} \left(1-\frac{z}{1+z}\right)^{N-1} \\ = {N+1\choose 2} \frac{1}{(1+z)^{N-1}}.$$

Substitute into the outer term to get

$$-\frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N-2} {N+1\choose 2} \frac{1}{(1+z)^{N-1}} \\ = -\frac{1}{2} [z^{N-1}] \frac{1}{1+z} = \frac{1}{2} (-1)^N.$$

The second piece is

$$\frac{1}{N(N+1)} [z^{N-2}] (1+z)^{N-2} [w^{N-1}] (1+w)^{N+2} \frac{1}{(1+zw/(1+z))^3}.$$

For this piece we obtain

$$\frac{1}{N(N+1)} [z^{N-2}] (1+z)^{N-2} \sum_{q=0}^{N-1} {N+2\choose N-1-q} (-1)^q {q+2\choose 2} \frac{z^q}{(1+z)^q}.$$

The remaining coefficient extractor cancels the term for $q=N-1:$

$$\frac{1}{N(N+1)} [z^{N-2}] (1+z)^{N-2} \sum_{q=0}^{N-2} {N+2\choose N-1-q} (-1)^q {q+2\choose 2} \frac{z^q}{(1+z)^q} \\ = \frac{1}{N(N+1)} \sum_{q=0}^{N-2} {N+2\choose N-1-q} (-1)^q {q+2\choose 2} \\ = - \frac{1}{N(N+1)} (-1)^{N-1} {N+1\choose 2} + \frac{1}{N(N+1)} \sum_{q=0}^{N-1} {N+2\choose N-1-q} (-1)^q {q+2\choose 2} \\ = \frac{1}{2} (-1)^N + \frac{1}{N(N+1)} \sum_{q=0}^{N-1} {N+2\choose N-1-q} (-1)^q {q+2\choose 2}.$$

Continuing, with the coefficient extractor enforcing the range,

$$\frac{1}{2} (-1)^N + \frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N+2} \sum_{q\ge 0} z^q (-1)^q {q+2\choose 2} \\ = \frac{1}{2} (-1)^N + \frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N+2} \frac{1}{(1+z)^3} \\ = \frac{1}{2} (-1)^N + \frac{1}{N(N+1)} [z^{N-1}] (1+z)^{N-1} \\ = \frac{1}{2} (-1)^N + \frac{1}{N(N+1)}.$$

Collecting the contributions from the two pieces we obtain at last

$$\bbox[5px,border:2px solid #00A000]{ (-1)^N + \frac{1}{N(N+1)}.}$$