I'm looking for a simple proof of the identity $$ \sum_{k=0}^n \frac{(-1)^k}{\binom{n}{k}} = \frac{n+1}{n+2} (1+(-1)^n) $$ relying only on elementary properties of binomial coefficients. I obtained this by starting with the integral representation $$ \frac{(a-1)!(b-1)!}{(a+b-1)!} = \int_0^{\infty} \frac{t^{b-1}}{(1+t)^{a+b}} \, dt, \hspace{0.5cm} a,b \in \mathbb{N}, $$ setting $a=n-k+1, b=k+1$ and adding, which gives $$ \frac{1}{n+1} \sum_{k=0}^n \frac{(-1)^k}{\binom{n}{k}} = \sum_{k=0}^n (-1)^k \frac{(n-k)! k!}{(n+1)!} = \int_0^{\infty} \frac{1}{(1+t)^{n+2}} \sum_{k=0}^n (-t)^k \, dt = \int_0^{\infty} \frac{1}{(1+t)^{n+2}} \frac{(-t)^{n+1}-1}{-t-1} \, dt $$ $$ = (-1)^n \int_0^{\infty} \frac{t^{n+1}}{(1+t)^{n+3}} \, dt + \int_0^{\infty} \frac{dt}{(1+t)^{n+3}} = \frac{(-1)^n+1}{n+2} $$ Is there an easier way, something which doesn't rely on the integral representation? Thanks.
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3See here. – Jean Marie Feb 28 '24 at 13:53
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For odd $n$, the identity follows immediately from $\binom{n}{k}=\binom{n}{n-k}$ because their reciprocals appear in pairs with opposite signs. – RobPratt Feb 29 '24 at 23:54
2 Answers
We have $$(-1)^kk!(n-k)!=\dfrac{1}{n+2}\left[(-1)^k(k+1)!(n-k)!-(-1)^{k-1}k!(n+1-k)!\right]$$ Therefore \begin{align*}\sum_{k=0}^n\dfrac{(-1)^k}{n\choose k}&=\dfrac{1}{n!}\sum_{k=0}^n(-1)^kk!(n-k)!\\ &=\dfrac{1}{n!(n+2)}\left[(-1)^n(n+1)!0!-(-1)0!(n+1)!\right] \\ &=\dfrac{(n+1)![(-1)^n+1]}{n!(n+2)} \\ & = \dfrac{(n+1)[(-1)^n+1]}{n+2}\end{align*}
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This answer is along the same lines as hxthanh's answer, but I had embellished with a bit more explanation, so I thought I would post it. I waited for a while so that their answer would get the attention it deserved; however, if it is decided that there is no added benefit to my version, I will remove it.
$$
\begin{align}
\frac{k!(n-k)!}{n!}&=\frac{n+1}{n+2}\left(\vphantom{\frac{k!}{1!}}\right.\overbrace{\frac{k!(n-k+1)!}{(n+1)!}}^{\frac{n-k+1}{n+1}\frac{k!(n-k)!}{n!}}+\overbrace{\frac{(k+1)!(n-k)!}{(n+1)!}}^{\frac{k+1}{n+1}\frac{k!(n-k)!}{n!}}\left.\vphantom{\frac{k!}{1!}}\right)\tag{1a}\\[3pt]
\frac{(-1)^k}{\binom{n}{k}}&=\frac{n+1}{n+2}\left(\frac{(-1)^k}{\binom{n+1}{k}}-\frac{(-1)^{k+1}}{\binom{n+1}{k+1}}\right)\tag{1b}\\
\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}&=\frac{n+1}{n+2}\left(1-(-1)^{n+1}\right)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a):}$ $\frac{n+1}{n+2}\left(\frac{n-k+1}{n+1}+\frac{k+1}{n+1}\right)=1$
$\text{(1b):}$ multiply by $(-1)^k$
$\text{(1c):}$ sum in $k$ from $0$ to $n$, taking advantage
$\phantom{\text{(1c):}}$ of the telescoping sum on the right hand side
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