Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$

Solution 1:

Remarking that \begin{equation} \sum_{k=1}^n\sin\left( \left( 2k-1 \right)x \right)=\frac{\sin^2nx}{\sin x} \end{equation} the proposed integral \begin{align} I_n&= \frac{2}{n}\int_0^{\frac{\pi }{2}}\frac{1-\cos (2 n x)}{\sin (x)} \, dx\\ &= \frac{4}{n}\int_0^{\frac{\pi }{2}}\frac{\sin^2nx}{\sin x},dx\\ &=\frac{4}{n}\sum_{k=1}^n\int_0^{\frac{\pi }{2}}\sin\left( \left( 2k-1 \right)x \right)\,dx\\ &=\frac{4}{n}\sum_{k=1}^n\frac{1}{2k-1}\\ &=\text{rhs} \end{align} which shows that the integral is equal to the rhs of the identity.

This decomposition suggests the use of the Chebyshev polynomials to evaluate the lhs, \begin{equation} \text{lhs}=\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!} \end{equation} Indeed, the Chebyshev polynomials of the first kind reads \begin{equation} T_n(z)=n\sum_{k=0}^n(-2)^k\frac{(k+n-1)!}{(n-k)!(2k)!}(1-z)^k \end{equation} and thus \begin{equation} \sum_{k=1}^n(-1)^{k-1}\frac{(n+k-1)!}{(n-k)!(2k)!}\left( 2(1-z) \right)^k=\frac{1}{n}\left( 1-T_n(z) \right) \end{equation} and with $Z=2(1-z)$, \begin{equation} \sum_{k=1}^n(-1)^{k-1}\frac{(n+k-1)!}{(n-k)!(2k)!}Z ^k=\frac{1}{n}\left[ 1-T_n(1-\frac{Z}{2}) \right] \end{equation} This summation is very similar to the proposed one. To introduce the missing factor $\frac{(k-1)!k!}{(2k)!}=\mathrm{B}(k,k+1)$ (here, $\mathrm{B}(k,k+1)$ is the Beta function), we use the integral representation: \begin{equation} \int_{0}^{\pi/2}{\sin^{2a-1}}\theta{\cos^{2b-1}}\theta\mathrm{d}\theta=\tfrac{1}{2}\mathrm{B}\left(a,b\right) \end{equation} with $a=k,b=k+1$, to express \begin{align} \mathrm{B}(k,k+1)&=2\int_{0}^{\pi/2}{\sin^{2k-1}}\theta{\cos^{2k+1}}\theta\,d\theta\\ &=2^{1-2k}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}\sin^{2k}2\theta\,d\theta \end{align} Thus \begin{align} \text{lhs}&=\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k+n-1)!}{(2 k)! (n-k)!}\mathrm{B}(k,k+1)\\ &=2\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}\,d\theta\sum _{k=1}^n \frac{(-1)^{k-1} (k+n-1)!}{(2 k)! (n-k)!}16^k2^{-2k}\sin^{2k}2\theta\\ &=\frac{2}{n}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta} \left[ 1-T_n(1-2\sin^22\theta) \right]\,d\theta\\ &=\frac{2}{n}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta} \left[ 1-T_n(\cos4\theta) \right]\,d\theta \end{align} But $T_n(\cos4\theta)=\cos 4n\theta$ and $1-\cos 4n\theta=2\sin^22n\theta$. We obtained then \begin{equation} \text{lhs}=\frac{4}{n}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}\sin^22n\theta\,d\theta\ \end{equation} By changing $\theta=u/2$ in the above integral and using simple trigonometric manipulations, \begin{align} \text{lhs}&=\frac{2}{n}\int_{0}^{\pi}\frac{\cos\frac{u}{2}}{\sin\frac{u}{2}}\sin^2nu\,du\\ &=\frac{4}{n}\int_{0}^{\pi}\frac{\cos^2\frac{u}{2}}{\sin u}\sin^2nu\,du\\ &=\frac{2}{n}\int_{0}^{\pi}\frac{\sin^2nu}{\sin u}\left( 1+ \cos u\right)\,du\\ &=\frac{2}{n}\int_{0}^{\pi}\frac{\sin^2nu}{\sin u}\,du+\frac{2}{n}\int_{0}^{\pi}\frac{\sin^2nu}{\sin u} \cos u\,du \end{align} By symmetry, the second integral vanishes and using symmetry for the first one, \begin{align} \text{lhs}&=\frac{4}{n}\int_{0}^{\pi/2}\frac{\sin^2nu}{\sin u}\,du\\ &=I_n \end{align}

Solution 2:

We show for $n\geq 1$ the validity of the equality chain : \begin{align*} \sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{2}{n}\int_0^{\frac{\pi }{2}} \frac{1-\cos (2 n z)}{\sin (z)} \, dz=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}\tag{1} \end{align*}

We start with the left-hand side of (1). We obtain for $n\geq 1$: \begin{align*} \color{blue}{\frac{2}{n}}&\color{blue}{\int_{0}^{\frac{\pi}{2}}\frac{1-\cos(2nz)}{\sin(z)}\,dz}\\ &=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\left(1-\sum_{j=0}^n(-1)^k\binom{2n}{2j}\cos^{2n-2j}(z)\sin^{2j}(z)\right)\frac{dz}{\sin(z)}\tag{2}\\ &=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\left(1-\sum_{j=0}^n(-1)^k\binom{2n}{2j}\left(1-\sin^2(z)\right)^{n-j}\sin^{2j}(z)\right)\frac{dz}{\sin(z)}\\ &=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\left(1-\sum_{j=0}^n(-1)^k\binom{2n}{2j}\sum_{k=0}^{n-j}\binom{n-j}{k}(-1)^k\sin^{2j+2k}(z)\right)\frac{dz}{\sin(z)}\\ &=\frac{2}{n}\sum_{j=0}^n\sum_{{k=0}\atop{(j,k)\ne(0,0)}}^{n-j}\binom{2n}{2j}\binom{n-j}{k}(-1)^{j+k-1}\int_{0}^{\frac{\pi}{2}}\sin^{2j+2k-1}(z)\,dz\\ &=\frac{2}{n}\sum_{j=0}^n\sum_{{k=0}\atop{(j,k)\ne(0,0)}}^{n-j}\binom{2n}{2j}\binom{n-j}{k}(-1)^{j+k-1}\frac{4^{j+k-1}}{2j+2k-1}\binom{2j+2k-2}{j+k-1}^{-1}\tag{3}\\ &=\frac{2}{n}\sum_{j=0}^n\sum_{{k=j}\atop{(j,k)\ne(0,0)}}^{n}\binom{2n}{2j}\binom{n-j}{k-j}(-1)^{k-1}\frac{4^{k-1}}{2k-1}\binom{2k-2}{k-1}^{-1}\tag{4}\\ &=\frac{2}{n}\sum_{k=1}^n\binom{2k-2}{k-1}^{-1}\frac{(-4)^{k-1}}{2k-1}\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}\tag{5}\\ &=\frac{2}{n}\sum_{k=1}^n\binom{2k-2}{k-1}^{-1}\frac{(-4)^{k-1}}{2k-1}\binom{n+k}{n-k}\frac{4^kn}{n+k}\tag{6}\\ &=\frac{1}{2}\sum_{k=1}^n\frac{(k-1)!(k-1)!}{(2k-2)!}\,\frac{(-1)^{k-1}16^k}{2k-1}\,\frac{(n+k)!}{(2k)!(n-k)!}\,\frac{1}{n+k}\\ &\,\,\color{blue}{=\sum_{k=1}^n\frac{(k-1)!k!}{(2k)!}\,\frac{(-1)^{k-1}16^k(n+k-1)!}{(2k)!(n-k)!}} \end{align*} and the claim follows.

Comment:

• In (2) we use the trigonometric summation identity \begin{align*} \cos(2nz)=\sum_{j=0}^n(-1)^k\binom{2n}{2j}\cos^{2n-2j}(z)\sin^{2j}(z) \end{align*}

• In (3) we use the identity $\int_{0}^{\frac{\pi}{2}}\sin^{2n+1}(z) dz=\frac{4^n}{2n+1}\binom{2n}{n}^{-1}$. See for instance this MSE post.

• In (4) we shift the index $k$ to start with $k=j$.

• In (5) we exchange the sums.

• In (6) we use the binomial identity $\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}=\binom{n+k}{n-k}\frac{4^kn}{n+k}$ valid for $1\leq k\leq n$. See for instance this MSE post.

The right-hand side of (1): We obtain \begin{align*} \color{blue}{\frac{2}{n}}&\color{blue}{\int_{0}^{\frac{\pi}{2}}\frac{1-\cos(2nz)}{\sin(z)}\,dz}\\ &=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\Re\left(2i\cdot\frac{1-e^{2inz}}{e^{iz}-e^{-iz}}\right)\,dz\tag{7}\\ &=-\frac{4}{n}\int_{0}^{\frac{\pi}{2}}\Re\left(ie^{iz}\cdot\frac{e^{2inz}-1}{e^{2iz}-1}\right)\,dz\\ &=-\frac{4}{n}\int_{0}^{\frac{\pi}{2}}\Re\left(ie^{iz}\sum_{k=0}^{n-1}e^{2ikz}\right)\,dz\tag{8}\\ &=-\frac{4}{n}\Re\left(i\sum_{k=0}^{n-1}\int_{0}^{\frac{\pi}{2}}e^{(2k+1)iz}\right)\,dz\\ &=-\frac{4}{n}\Re\left(\left.\sum_{k=0}^{n-1}\frac{1}{2k+1}e^{(2k+1)iz}\right|_{0}^{\frac{\pi}{2}}\right)\\ &=\frac{4}{n}\sum_{k=0}^{n-1}\frac{1}{2k+1}\\ &\,\,\color{blue}{=\frac{4}{n}\sum_{k=1}^{n}\frac{1}{2k-1}}\tag{9} \end{align*} and the claim follows.

Comment:

• In (7) we use the identities $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$.

• In (8) we use the finite geometric summation formula.

• In (9) we shift the index to start with $k=1$.

Note: When using binomial coefficients and the harmonic numbers $H_n=\sum_{k=1}^n\frac{1}{k}$ for $n\geq 1$ and $H_0=0$ we can write the equality chain (1) for $n\geq 1$ as \begin{align*} -\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{n-k}\binom{2k}{k}^{-1} =\frac{2}{n}\int_0^{\frac{\pi }{2}} \frac{1-\cos (2 n z)}{\sin (z)} \, dz =\frac{4}{n}\left(H_{2n-1}-\frac{1}{2}H_{n-1}\right) \end{align*}

Solution 3:

Unfinished approach that is too long for a comment:

I tried to use Sister Celine's method but there are annoying details:

Let $$F(n,k)=\frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}.$$

Then whenever $F(n,k)\neq0$, $$F(n+1,k)/F(n,k)=\frac{k+n}{1-k+n}$$ and $$F(n,k+1)/F(n,k)=-\frac{4 k (n-k) (k+n)}{(k+1) (2 k+1)^2},$$ so by Sister Celine's method we find that $F$ satisfies the recursion

\begin{equation} \sum_{r=0}^3\sum_{s=0}^1 a_{r,s}(n) F(n-r,k-s)=0 \end{equation}

where the $a_{r,s}(n)$ equal

$$\left( \begin{array}{cc} (1-2 n)^2 (n-2) n & 0 \\ -(n-1) (2 n-1) (n (6 n-17)+9) & 8 (n-2) (n-1)^2 (2 n-1) \\ (n-2) (2 n-1) (n (6 n-19)+12) & -8 (n-2)^2 (n-1) (2 n-1) \\ -(n-3) (n-1) (2 n-5) (2 n-1) & 0 \\ \end{array} \right)$$

whenever all $F(n-r,k-s)$ are defined. Now we would like to use this in order to deduce a recurrence for the sum $$G(n)=\sum_{k=1}^n F(n,k),$$ however we get issues since $F(n,0)$ is not well-defined. So maybe studying $\sum_{k=2}^n F(n,k)$ works better.

In fact, we (mysteriously) get the following recurrence that I don't have the time to figure out a proof for:

$$\left(-2 n^3+13 n^2-26 n+15\right) G(n-3)+\left(-2 n^3+9 n^2-14 n+8\right) G(n-2)+\left(2 n^3-9 n^2+14 n-7\right) G(n-1)+\left(2 n^3-5 n^2+2 n\right) G(n)=16 n-24.$$

Now we would have to prove that the right-hand side also satisfies this recurrence and we would be done.