Another binomial coefficients sum: $\sum_{k=0}^{n-1} {n-k\choose k}\frac {(-1)^k}{n-k}$

Question

Is there a closed form of the sum $$\sum_{k=0}^{n-1} {n-k\choose k}\frac {(-1)^k}{n-k}$$?

I cannot even guess what the sum will be. (We assume, here, that $i\choose j$=0 if $i < j$.)

Answer 1

The following is valid for $n\geq 1$

\begin{align*} \sum_{k=0}^{n-1} {n-k\choose k}\frac {(-1)^k}{n-k}= \begin{cases} \frac{2}{n}(-1)^n&n\equiv 0 \pmod{3}\\ \frac{1}{n}(-1)^{n-1}&otherwise \end{cases}\tag{1} \end{align*}

We will show the validity of (1) in two steps. At first we derive a generating function for the binomial expression. In the second step we extract the coefficients and show they correspond with the RHS of (1).

Step 1: Generating functions

The binomial expression in (1) can be written as

\begin{align*} \binom{n-k}{k}\frac{1}{n-k}=\frac{1}{n}\left(\binom{n-k}{k}+\binom{n-k-1}{k-1}\right) \end{align*}

and we derive two generating functions for the binomial expressions on the RHS.

Let's consider the formal power series \begin{align*} A(t)=\sum_{k=0}^{\infty}a_kt^k \end{align*}

Similar to Euler's series transformation formula

\begin{align*} \frac{1}{1-t}A\left(\frac{t}{1-t}\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}a_k\right)t^n \end{align*}

which provides us with a generating function for $\sum_{k=0}^n\binom{n}{k}a_k$, we consider the transformation $\frac{1}{1-t}A\left(\frac{t^2}{1-t}\right)$.

We obtain

\begin{align*} \frac{1}{1-t}A\left(\frac{t^2}{1-t}\right)&=\sum_{k=0}^{\infty}a_kt^{2k}(1-t)^{-(k+1)}\\ &=\sum_{k=0}^{\infty}a_kt^{2k}\sum_{l=0}^{\infty}\binom{-(k+1)}{l}(-t)^l\tag{2}\\ &=\sum_{k=0}^{\infty}a_kt^{2k}\sum_{l=0}^{\infty}\binom{k+l}{l}(-t)^l\tag{3}\\ &=\sum_{n=0}^{\infty}\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}\binom{k+l}{l}a_k\right)t^n\tag{4}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n-k}{n-2k}a_k\right)t^n\tag{5}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n-k}{k}a_k\right)t^n\tag{6}\\ \end{align*}

Comment:

• In (2) we use the series expansion of the binomial series

• In (3) we use the identity $\binom{r}{s}=\binom{-r+s-1}{s}(-1)^s$

• In (4) we multiply the series

• In (5) we replace $l$ with $n-2k$.

Next we obtain a generating function $B(t)$ by setting $a_k=(-1)^k$ in $A(t)$

\begin{align*} B(t)=\frac{1}{1+t}=\sum_{n=0}^{\infty}(-1)^nt^n \end{align*}

Since \begin{align*} \frac{1}{1-t}B\left(\frac{t^2}{1-t}\right)=\frac{1}{1-t}\frac{1}{1+\frac{t^2}{1-t}}=\frac{1}{1-t+t^2} \end{align*}

we get from (6) the following identity \begin{align*} \sum_{n=0}^{\infty}\left(\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n-k}{k}(-1)^k\right)t^n =\frac{1}{1-t+t^2}\tag{7} \end{align*}

In an analogous manner we obtain

\begin{align*} A\left(\frac{t^2}{1-t}\right)&=a_0+\sum_{k=1}^\infty a_k\left(\frac{t^2}{1-t}\right)^k\\ &=\cdots\\ &=a_0+\sum_{n=1}^{\infty}\left(\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\binom{n-k-1}{k-1}a_k\right)t^n \end{align*}

and get

\begin{align*} B\left(\frac{t^2}{1-t}\right)&=\frac{1}{1+\frac{t^2}{1-t}}=\frac{1-t}{1-t+t^2}\\ &=1+\sum_{n=1}^{\infty}\left(\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\binom{n-k-1}{k-1}(-1)^k\right)t^n\tag{8} \end{align*}

In the following we use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ of the series.

We can now put (7) and (8) together and derive

\begin{align*} \sum_{k=0}^{n-1} &{n-k\choose k}\frac {(-1)^k}{n-k}=\frac{1}{n}[t^n]\left(\frac{1}{1-t}B\left(\frac{t^2}{1-t}\right)+B\left(\frac{t^2}{1-t}\right)\right)\\ &=\frac{1}{n}[t^n]\left(\frac{1}{1-t+t^2}+\frac{1-t}{1-t+t^2}\right)\\ &=\frac{1}{n}[t^n]\frac{2-t}{1-t+t^2}\tag{9}\\ \end{align*}

$$ $$

Step 2: Coefficient extraction

In order to find the coefficient of (9) we consider at first $\frac{1}{1-t+t^2}$ and show

The following is valid \begin{align*} [t^n]\frac{1}{1-t+t^2}=\frac{2}{\sqrt{3}}\sin\left((n+1)\frac{\pi}{3}\right)\qquad\qquad n\geq 0\tag{10} \end{align*}

Let $t_0,t_1$ denote the zeros of the polynomial $1-t+t^2$. Partial fraction decomposition gives \begin{align*} \frac{1}{(t-t_0)(t-t_1)}&=\frac{1}{t_0-t_1}\left(\frac{1}{t-t_1}-\frac{1}{t-t_0}\right)\\ &=\frac{1}{t_0-t_1}\sum_{k=0}^{\infty}\left(\frac{1}{{t_0}^{k+1}}-\frac{1}{t_1^{k+1}}\right)t^k \end{align*} Since \begin{align*} t_{0,1}=\frac{1}{2}\left(1\pm i\sqrt{3}\right)=e^{\pm i\frac{\pi}{3}} \end{align*} and $t_0-t_1=i\sqrt{3}$ we obtain for $n\geq 0$ \begin{align*} [t^n]\frac{1}{(t-t_0)(t-t_1)}&=\frac{1}{{(t_0t_1)}^{n+1}}\frac{t_0^{n+1}-t_1^{n+1}}{t_0-t_1}\\ &=\frac{e^{i(n+1)\frac{\pi}{3}}-e^{-i(n+1)\frac{\pi}{3}}}{i\sqrt{3}}\\ &=\frac{2}{\sqrt{3}}\sin\left((n+1)\frac{\pi}{3}\right) \end{align*} and the claim (10) follows.

Since we need the coefficient $[t^n]\frac{2-t}{1-t+t^2}$ we show the following is valid \begin{align*} [t^n]\frac{2-t}{1-t+t^2}=2\cos\left(n\frac{\pi}{3}\right)\qquad\qquad n\geq 0\tag{11} \end{align*}

Using (10) we obtain

\begin{align*} [t^n]\frac{2-t}{1-t+t^2}&=\frac{4}{\sqrt{3}}\sin\left((n+1)\frac{\pi}{3}\right) -\frac{2}{\sqrt{3}}\sin\left(n\frac{\pi}{3}\right)\\ &=\frac{2e^{i(n+1)\frac{\pi}{3}}-2e^{-i(n+1)\frac{\pi}{3}}}{i\sqrt{3}}- \frac{e^{in\frac{\pi}{3}}-e^{-in\frac{\pi}{3}}}{i\sqrt{3}}\\ &=-\frac{i}{\sqrt{3}}\left(e^{in\frac{\pi}{3}}\left(2e^{i\frac{\pi}{3}}-1\right) -e^{-in\frac{\pi}{3}}\left(2e^{-i\frac{\pi}{3}}-1\right)\right)\\ &=-\frac{i}{\sqrt{3}}\left(e^{in\frac{\pi}{3}}i\sqrt{3} -e^{-in\frac{\pi}{3}}(-i\sqrt{3})\right)\\ &=e^{in\frac{\pi}{3}}+e^{-in\frac{\pi}{3}}\\ &=2\cos\left(n\frac{\pi}{3}\right) \end{align*} and the claim (11) follows.

We conclude from (9) and (11)

\begin{align*} \sum_{k=0}^{n-1} &{n-k\choose k}\frac {(-1)^k}{n-k}=\frac{2}{n}\cos\left(n\frac{\pi}{3}\right)\qquad\qquad n\geq 1 \end{align*}

Finally, note that \begin{align*} \frac{2}{n}\cos\left(n\frac{\pi}{3}\right)= \begin{cases} \frac{2}{n}(-1)^n&n\equiv 0 \pmod{3}\\ \frac{1}{n}(-1)^{n-1}&otherwise \end{cases} \end{align*} which proves that (1) is valid.