I am trying to prove (or disprove) the non-vanishing of the finite summation $$\sum_{k=0}^n (-1)^k \frac{2k+1}{\sqrt{(n - k)!(n+1+k)!}} \neq 0 \quad \forall n < \infty.$$
Alternatively, that the finite partial sums in the infinite series $$\sum_{k=0}^\infty (-1)^k \frac{2k+1}{\sqrt{(n - k)!(n+1+k)!}} = 0$$ are not zero. i.e. they do not miraculously cancel out before they reach infinity.
I strongly suspect the above to be true.
Attempts
(note this is not my area of expertise).
Here is a numerical plot of the partial sums up to $n=15$. It appears to be an absolutely convergent series, with even $n$ being positive-definite and odd $n$ being negative-definite. Not sure how to prove this in general.
Given a fixed $n$, I was hoping the absolute value of each term $a_k = \frac{2k+1}{\sqrt{(n - k)!(n+1+k)!}}$in $S_n$ would be monotonically decreasing, showing for example that an even $n$ would always be a bit more positive than negative. But a numerical plot shows this is not the case: coefficient for n=15. I still tried to see if anything can be gained by somehow identifying when in k the peak occurs, maybe to split the sum into before and after.
I have calculated that $$ a_{k+1} = \sigma_k a_k , \qquad \, \sigma_k = \sqrt{\frac{n-k}{n+k+2}} \Big(\frac{2k+3}{2k+1}\Big).$$ And so the summation can be re-expressed as $$ \sum_{k=0}^n (-1)^k \Big[ \prod_{j=0}^k \sigma_j \Big]a_0, \qquad a_0 = \frac{1}{\sqrt{n! (n+1)!}} .$$
The product can be calculated as $$ \prod_{j=0}^k \sigma_j = \frac{a_k}{a_0} = \sqrt{\frac{k!(n+1)!}{(n-k)!(n+1+k)!}} = \sqrt{\frac{{}_nP_k}{{}_{n+1+k}P_k}} $$
Now I guess I would take the derivative of this with respect to $k$ (using the Gamma function) and set to zero. But that looks too complicated for me. At this point I'm not sure how to proceed (and/or I have made a mistake, and if this is the even right approach). Hopefully there is a simple argument I'm missing. Maybe something to do with the denominators and not being able to cancel out if there is no gcd?
UPDATE
I have found the difference between two absolute terms in a given partial sum $S_n$ to be $$ a_{k+1} - a_k = \Big( \frac{2k+3}{2k+1} \sqrt{\frac{n-k}{n+k+2}} - 1 \Big)a_k $$ A monotonically decreasing sequence will have $$ a_{k+1} - a_k < 0 \qquad \Rightarrow \qquad \frac{2k+3}{2k+1} \sqrt{\frac{n-k}{n+k+2}} < 1. $$ With $k,n \geq 0$ in mind, this inequality may be simplified to $$ n \leq k^2 + 2k + \frac{1}{4}$$ and inverted to $$ k \geq \sqrt{n + \frac{3}{4}} - 1.$$ This is saying that any fixed-$n$ partial sum $S_n$ will eventually reach its total via an alternating sum that absolutely decreases monotonically. For example see this plot for the n=60 case; the blue line is the RHS of the above inequality.
And by flipping the above inequalities it's clear that the first $\approx \sqrt{n}$ elements are monotonically increasing. So for a fixed $n$, the partial sum $S_n$ can be broken down: sum up the first $\approx \sqrt{n}$ elements and get some number. The remaining $n - \sqrt{n}$ subsequent additions lead to a decreasingly small bobbing up and down around that number.
This feels like progress but it's not complete.