# A closed form for $\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$

Mathematica does give an analytic form for $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}.$$ The question here is: How to find a simpler closed form for this alternating summation by hand. The summation of the absolute seies has been discussed earlier in MSE: Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$

Since

$$\frac{\binom nk}{k+1}=\frac{\binom{n+1}{k+1}}{n+1}\;,$$

we have

$$\begin{eqnarray} \sum_{k=0}^n\frac{(-1)^k\binom nk^2}{(k+1)^2} &=& \frac1{(n+1)^2}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}^2 \\ &=& \frac1{(n+1)^2}\left(1-\sum_{k=0}^{n+1}(-1)^k\binom{n+1}k^2\right)\;. \end{eqnarray}$$

With

$$\sum_{k=0}^nq^k\binom nk^2=(1-q)^nP_n\left(\frac{1+q}{1-q}\right)\;,$$

where $$P_n$$ is the $$n$$-th Legendre polynomial, this is

$$\frac1{(n+1)^2}\left(1-2^{n+1}P_{n+1}(0)\right)\;,$$

where

$$P_l(0)= \begin{cases} \frac{(-1)^m}{4^m}\binom{2m}m&l=2m\\0&l=2m+1 \end{cases}$$

(see Wikipedia).

The factor $$4^m$$ cancels, so the result is

$$\sum_{k=0}^n\frac{(-1)^k\binom nk^2}{(k+1)^2}= \begin{cases} \frac{1-(-1)^m\binom{2m}m}{(2m)^2}&n=2m-1\;,\\ \frac{1}{(2m+1)^2}&n=2m\;. \end{cases}$$

• When $n$ is odd your answer is right but when $n$ is even please check $S_n=\frac{1}{(n+1)^2}.$ Mar 10, 2020 at 18:20
• @DrZafarAhmedDSc: Thanks; apparently Gary already fixed the mistake I'd made in substituting $P_l(0)$ into the final result. Mar 10, 2020 at 20:18

Use Binomial identity: $$(1+t)^n=\sum_{k=0}^{n} {n \choose k}t^n. \tag{1}$$ Integration of $$(1)$$ from $$t=0$$ to $$t=x$$ gives $$\frac{(1+x)^{n+1}-1}{n+1}= \sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}.\tag{2}$$ We can change $$x$$ to $$-1/x$$ in $$(2)$$ to get $$\frac{(1-1/x)^{n+1}-1}{n+1}= \sum_{k=0}^n (-1)^k {n \choose k}\frac{x^{-k-1}}{k+1} . \tag{3}$$ Multiplying $$(2)$$ and $$(3)$$ and collecting terms free of $$x$$ on RHS, we get $$\frac{x^{-n-1}\big[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}\big]}{(n+1)^2}=\sum_{k=0}^{n}\frac{ (-1)^k {n \choose k}^2}{(k+1)^2} x^0+\dots$$ Equating the coefficients yields $$S_n = \sum_{j=0}^{n} \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$$ $$S_N=[x^{n+1}]~\frac{[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2}$$ $$S_n=\frac{-(-1)^{(n+1)/2}{n+1 \choose (n+1)/2}+1}{(n+1)^2}, \text{when n is odd}$$ $$S_n= \frac{1}{(n+1)^2}, \text{when n is even}$$