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 $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$


2 Answers 2



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

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}


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

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

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


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

(see Wikipedia).

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

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

  • $\begingroup$ When $n$ is odd your answer is right but when $n$ is even please check $S_n=\frac{1}{(n+1)^2}.$ $\endgroup$
    – Z Ahmed
    Mar 10, 2020 at 18:20
  • 1
    $\begingroup$ @DrZafarAhmedDSc: Thanks; apparently Gary already fixed the mistake I'd made in substituting $P_l(0)$ into the final result. $\endgroup$
    – joriki
    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 $$S_n = \sum_{j=0}^{n} \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$$ $$S_N=[x^{n+1}]~\frac{[(-1)^{n+1}(1-x^2)^{n+1}-(x-1)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2} $$ $$S_n=\frac{-(-1)^{(n+1)/2}{n+1 \choose (n+1)/2}+1}{(n+1)^2}, \text{when $n$ is odd}$$ $$S_n= \frac{1}{(n+1)^2}, \text{when $n$ is even} $$


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