How to prove $ \sum_{k=0}^n \frac{(-1)^{n+k}{n+k\choose n-k}}{2k+1}=\frac{-2\cos\left(\frac{2(n-1)\pi}{3}\right)}{2n+1}$

How to prove

$$\sum_{k=0}^n \binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=-\frac{2}{2n+1}\,\cos\left(\frac{2(n-1)\pi}{3}\right)\;\text{?}$$

I have a proof by induction for it, but it isn't simple! I want to seek an interesting proof for it. (Sorry for my English study is very bad) Can you help me re-open this question? Thank you very much!

Solution 1:

We may exploit: $$ F_{2n+1}(x) = \sum_{k=0}^{n}\binom{n+k}{n-k}x^{2k}\tag{1} $$ where $F_n(x)$ is a Fibonacci polynomial. That gives:

$$\sum_{k=0}^{n}\binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=(-1)^n\int_{0}^{1}\frac{\alpha(x)^{2n+1}-\beta(x)^{2n+1}}{\alpha(x)-\beta(x)}\,dx,\tag{2}$$ where: $$ \alpha(x) = \frac{ix+\sqrt{4-x^2}}{2},\qquad \beta(x) = \frac{ix-\sqrt{4-x^2}}{2},\tag{3}$$ so we just need to use the substitution $x=2\sin\theta$ in the RHS of $(2)$ to prove our claim, since the RHS of $(2)$ is so converted into: $$ 2(-1)^n \int_{0}^{\frac{\pi}{6}}\cos((2n+1)\theta)\,d\theta.\tag{4}$$ As pointed by r9m, the same can be achieved by using Chebyshev polynomials of the second kind in the first place.

Solution 2:

Let's try a generating function approach.

Multiply by $x^ky^n$ and sum in both $k$ and $n$: $$ \begin{align} \sum_{k,n}\binom{n+k}{n-k}x^ky^n &=\sum_{k,n}\binom{n}{2k}\frac{x^k}{y^k}y^n\tag{1}\\ &=\frac1{1-y}\sum_k\frac{x^ky^k}{(1-y)^{2k}}\tag{2}\\ &=\frac{1-y}{(1-y)^2-xy}\tag{3} \end{align} $$ Explanation:
$(1)$: $\binom{n+k}{n-k}=\binom{n+k}{2k}$, then substitute $n\mapsto n-k$
$(2)$: $\sum\limits_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$
$(3)$: $\sum\limits_kx^k=\frac1{1-x}$

Substitute $x\mapsto-x^2$, $y\mapsto y^2$ and integrate $(3)$ in $x$ from $0$ to $1$: $$ \begin{align} \sum_{k,n}\frac{(-1)^k}{2k+1}\binom{n+k}{n-k}y^{2n} &=\int_0^1\frac{1-y^2}{(1-y^2)^2+x^2y^2}\,\mathrm{d}x\tag{4}\\ &=\frac1y\,\tan^{-1}\!\left(\frac{y}{1-y^2}\right)\tag{5}\\[6pt] &=\frac1y\left[\tan^{-1}\left(e^{i\pi/3}y\right)+\tan^{-1}\left(e^{-i\pi/3}y\right)\right]\tag{6}\\ &=2\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\cos\left(\frac{(2n+1)\pi}3\right)y^{2n}\tag{7} \end{align} $$ Explanation:
$(4)$: substitute $x\mapsto-x^2$, $y\mapsto y^2$ and integrate $(3)$ in $x$ from $0$ to $1$
$(5)$: perform the integration
$(6)$: use the formula for tangent of a sum
$(7)$: use the series for arctangent

Equation $(7)$ implies that $$ \begin{align} \sum_{k=0}^n\frac{(-1)^{n+k}}{2k+1}\binom{n+k}{n-k} &=\frac2{2n+1}\cos\left(\frac{(2n+1)\pi}3\right)\tag{8}\\ &=-\frac2{2n+1}\cos\left(\frac{(2n-2)\pi}3\right)\tag{9} \end{align} $$ Explanation:
$(8)$: equate identical powers of $y$ in $(7)$
$(9)$: $\cos(x+\pi)=-\cos(x)$

Solution 3:

My Solution

We have: \begin{align*}\sum_{k=0}^n\dfrac{(-1)^{n+k}}{2k+1}{n+k\choose n-k}&=\sum_{k=0}^n \dfrac{(-1)^{n+k}(n+k)!}{(2k+1)!(n-k)!} \\ &=\dfrac{1}{2n+1}\sum_{k=0}^n\dfrac{(-1)^{n+k}(n+k)!\left[(n+k+1)+(n-k)\right]}{(2k+1)!(n-k)!}\\&=\dfrac{1}{2n+1}\sum_{k=0}^n\left[(-1)^{n+k}{n+1+k\choose 2k+1} -(-1)^{n-1+k}{n+k\choose 2k+1}\right]\\ &=\dfrac{1}{2n+1}\left(S_n-S_{n-1}\right)\end{align*} Where: \begin{align*}S_n&=\sum_{k=0}^n(-1)^{n+k}{n+1+k\choose 2k+1}=\sum_{k=0}^n(-1)^{n+k}{n+1+k\choose n-k}\\ &=\sum_{k=0}^n(-1)^k{2n+1-k\choose k}\qquad\text{(reverse index)}\end{align*} So, easy proof $\quad S_n=-S_{n-1}-S_{n-2}\quad$ (reference here)

$\{S_n\}_{n\ge 0}\;:\;\{1,-1,0,1,-1,0,1,-1,0,...\}$

$\{S_n-S_{n-1}\}_{n\ge 1}\;:\;\{-2,1,1,-2,1,1,-2,1,1,...\}$

DONE! :D

Solution 4:

This is slightly indirect way of approaching the problem but just for sake of variety I am adding a new approach.

Making the substitution $k \mapsto n-j$,

$$\sum_{k=0}^{n} (-1)^{n-k}\binom{n+k}{n-k}x^{2k} = \sum_{j = 0}^{n} (-1)^{j}\binom{2n-j}{j}x^{2n-2j} = U_{2n}\left(\frac{x}{2}\right)$$

where, $U_{n}(x)$ is the Chebyshev Polynomial of the second kind. See here for an elementary proof of the rightmost equality.

Making the substitution $x = 2\cos \theta$, we may write the above expresion as:

$$U_{2n}\left(\frac{2\cos \theta}{2}\right) = \frac{\sin (2n+1)\theta}{\sin \theta}$$

Thus, $$\begin{align}\sum_{k=0}^{n} \frac{(-1)^{n-k}}{2k+1}\binom{n+k}{n-k} &= \int_0^1 \sum_{k=0}^{n} (-1)^{n-k}\binom{n+k}{n-k}x^{2k} \,\mathrm{d}x \\ &= \int_0^1 U_{2n}(x/2)\,\mathrm{d}x\\&= 2\int_{\pi/3}^{\pi/2} \sin (2n+1)\theta \,\mathrm{d}\theta \\ &= \left.-2\frac{\cos (2n+1)\theta}{2n+1}\right\vert_{\pi/3}^{\pi/2}\\ &= \frac{2}{2n+1}\cos \left(\frac{(2n+1)\pi}{3}\right)\end{align}$$

as required.