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Find a closed form expression for $$\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}$$

where $n$ is an even positive integer.

I tried using binomial identities, but since the binomial coefficient is in the denominator, I can't do much. Also, I'm seeking a non-calculus, elementary solution, if possible.

Thanks.

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  • $\begingroup$ you should get $\frac{2(n+1)!}{(n+2)n!}$. $\endgroup$
    – Math-fun
    Aug 19, 2015 at 16:26
  • $\begingroup$ I have provided an alternate and shorter approach to your question. $\endgroup$
    – MathGod
    Aug 20, 2015 at 13:55
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    $\begingroup$ @Ishu Thanks a lot! That's an extremely elegant approach! $\endgroup$
    – user196761
    Aug 20, 2015 at 14:43

5 Answers 5

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$$\underline{\text{Method 1}}$$

$$\text{S}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}$$

$$=\displaystyle\sum_{r=0}^{n}(-1)^r\dfrac{r!(n-r)!}{n!}$$

$$=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{blue}{(n-r+1+r+1)}}{n!}\color{red}{\times\dfrac{1}{(n+2)}}$$

$$=\displaystyle\sum_{r=0}^{n}\dfrac{1}{n!(n+2)}\left[(-1)^r\left\{r!(n-r+1)!+(n-r)!(r+1)!\right\}\right]$$

$$\color{blue}{=\displaystyle\sum_{r=0}^{n}\dfrac{1}{n!(n+2)}\left[\left\{T_{r} - T_{r+1}\right\}\right]}\tag{*}$$

where $T_{r}= (-1)^r r!(n-r+1)!$

Clearly, $(*)$ is a telescoping series. Evaluating it, we have,

$$\text{S}=(-1)^nT_{n+1}+T_{0}$$

$$=(-1)^n\dfrac{(n+1)!}{n!(n+2)} + \dfrac{(n+1)!}{n!(n+2)}$$

Since $n$ is even,

$$\color{red}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}$$


$$\underline{\text{Method 2}}$$

Consider the following Lemma.

Lemma : $$\color{brown}{\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}=0}$$

where $k$ is an odd positive integer.

Proof : $$\text{S}=\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}$$

$$=\displaystyle \sum_{r=0}^k\dfrac{(-1)^{k-r}}{\dbinom{k}{k-r}}$$

$$\displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r) = \sum_{r=0}^{n} f(n-r) \right)} $$

$$= -\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}$$

$$\implies \text{S}=-\text{S}$$

$$\implies \text{S}=0$$

This completes the proof of the Lemma.

Now, since $n$ is even, $n+1$ is odd.

Using the Lemma, we have,

$$\color{blue}{\displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}=0} \tag{1}$$

Let,

$$\color{red}{\text{J}=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}}}\tag{2}$$

Operating $(1)+(2)$,

$$\implies \text{J}+0=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}$$

$$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n+1}{r}} +\dfrac{(-1)}{\dbinom{n+1}{n+1}}$$

$$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{n-r}}{\dbinom{n+1}{n-r}} -1 \quad \displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r)=\sum_{r=0}^{n} f(n-r)\right)}$$

$$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n+1}{r+1}} -1$$

$$=\displaystyle \sum_{r=0}^n(-1)^r\left\{\dfrac{1}{\dbinom{n}{r}} + \dfrac{r+1}{n+1} \cdot \dfrac{1}{\dbinom{n}{r}}\right\} -1 \quad \displaystyle\color{red}{\left[\because \dbinom{n}{r}=\dfrac{n}{r}\dbinom{n-1}{r-1} \ \text{&} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]}$$

$$\implies \text{J} = \displaystyle\left(\dfrac{n+2}{n+1}\right) \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}+ \dfrac{1}{n+1} \sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}} -1$$

$$\implies \text{J}+1 = \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{1}{n+1} \underbrace{\color{#66f}{\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}}}_{\huge{= \ \text{G} \ (\text{let})}} $$

$$\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{G}}{n+1} \tag{3} $$

Now,

$$\text{G}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}$$

$$=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{n-r}\cdot (n-r)}{\dbinom{n}{n-r}}$$

$$=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot (n-r)}{\dbinom{n}{r}}$$

$$=\displaystyle n\sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n}{r}} - \sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot r}{\dbinom{n}{r}}$$

$$\implies \text{G}=n\cdot \text{J}-\text{G}$$

$$\implies\color{red}{\text{G}=\dfrac{\text{J}n}{2}} \tag{4}$$

From $(3)\ \text{&} \ (4)$,

$$\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{J}n}{2(n+1)}$$

$$\color{green}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square$$

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    $\begingroup$ Awesome! Both your methods are extremely ingenious! Thanks. $\endgroup$
    – user196761
    Aug 20, 2015 at 14:27
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    $\begingroup$ Now that I've seen Method 1, I would vote for this answer again if I could. $\endgroup$
    – user84413
    Aug 23, 2015 at 23:18
  • $\begingroup$ Haha how were you even able to come up with this AND type it up all in under one hour? $\endgroup$
    – Ovi
    Jul 4, 2016 at 3:34
  • $\begingroup$ @Ovi I initially typed only Method $2$. Method $1$ (telescoping) was added later. $\endgroup$
    – MathGod
    Jul 4, 2016 at 3:37
  • $\begingroup$ Would you mind sharing how you came up with these solutions? Do you just have a lotof experience with algebraic sums and recognized things quickly? $\endgroup$
    – Ovi
    Oct 28, 2017 at 18:29
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Brute force with the Beta function: $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}} &=\sum_{k=0}^n(-1)^k\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+1)}\tag{1a}\\ &=(n+1)\sum_{k=0}^n(-1)^k\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}\tag{1b}\\ &=(n+1)\sum_{k=0}^n(-1)^k\operatorname{B}(k+1,n-k+1)\tag{1c}\\ &=(n+1)\int_0^1\sum_{k=0}^n(-1)^kt^k(1-t)^{n-k}\,\mathrm{d}t\tag{1d}\\ &=(n+1)\int_0^1(1-t)^n\frac{1+\left(\frac t{1-t}\right)^{n+1}}{1+\frac{t}{1-t}}\,\mathrm{d}t\tag{1e}\\ &=(n+1)\int_0^1\left[(1-t)^{n+1}+t^{n+1}\right]\,\mathrm{d}t\tag{1f}\\ &=\bbox[5px,border:2px solid #C0A000]{2\frac{n+1}{n+2}}\tag{1g} \end{align} $$ Explanation:
$\text{(1a)}$: $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$ and $n!=\Gamma(n+1)$
$\text{(1b)}$: $\Gamma(n+2)=(n+1)\Gamma(n+1)$
$\text{(1c)}$: definition of the Beta function in terms of the Gamma function
$\text{(1d)}$: definition of the Beta function in terms of an integral
$\text{(1e)}$: $\frac{1+x^{n+1}}{1+x}=1-x+x^2-\dots+x^n$ when $n$ is even
$\text{(1f)}$: simplify the integrand
$\text{(1g)}$: integrate


A more elementary approach:

For $k\gt0$, partial fractions gives $$ \frac1{\binom{n}{k}}=\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}\frac j{n-j+1}\tag{2} $$ Therefore, $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}} &=1+\sum_{k=1}^n\frac{(-1)^k}{\binom{n}{k}}\tag{3a}\\ &=1+\sum_{k=1}^n\sum_{j=1}^k(-1)^j\binom{k}{j}\frac j{n-j+1}\tag{3b}\\ &=1+\sum_{j=1}^n\sum_{k=j}^n(-1)^j\binom{k}{j}\frac j{n-j+1}\tag{3c}\\ &=1+\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac j{n-j+1}\tag{3d}\\ &=1+\sum_{j=0}^n(-1)^j\binom{n}{j}\frac{(n+1)\,j}{(j+1)(n-j+1)}\tag{3e}\\ &=1+\frac{n+1}{n+2}\sum_{j=0}^n(-1)^j\binom{n}{j}\left(\frac j{j+1}+\frac j{n-j+1}\right)\tag{3f}\\ &=1+\frac{n+1}{n+2}\sum_{j=0}^n(-1)^j\binom{n}{j}\frac n{j+1}\tag{3g}\\ &=1+\frac{n+1}{n+2}\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac n{n+1}\tag{3h}\\ &=1+\frac n{n+2}\tag{3i}\\ &=\bbox[5px,border:2px solid #C0A000]{2\frac{n+1}{n+2}}\tag{3j} \end{align} $$ Explanation:
$\text{(3a)}$: separate the $k=0$ term
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: change order of summation
$\text{(3d)}$: $\sum\limits_{k=j}^n\binom{k}{j}=\binom{n+1}{j+1}$
$\text{(3e)}$: $\binom{n+1}{j+1}=\binom{n}{j}\frac{n+1}{j+1}$
$\text{(3f)}$: partial fractions
$\text{(3g)}$: substitute $j\mapsto n-j$ in the right summand
$\text{(3h)}$: $\binom{n}{j}\frac1{j+1}=\binom{n+1}{j+1}\frac1{n+1}$
$\text{(3i)}$: $\sum\limits_{j=0}^n(-1)^j\binom{n+1}{j+1}=1$
$\text{(3j)}$: simplify

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For a fixed $k$, if we regard $\frac{1}{\binom{n}{k}}$ as a meromorphic function of $n$, we have:

$$ \text{Res}\left(\binom{n}{k}^{-1},\,n=r\right)=k(-1)^{k+r+1}\binom{n-1}{r} $$ for every $r\in[0,k-1]$. That implies: $$ \text{Res}\left(\sum_{k=0}^{n}\frac{(-1)^k}{\binom{n}{k}},n=0\right)=0-1-2-\ldots-n=-\binom{n+1}{2}$$ as well as: $$ \text{Res}\left(\sum_{k=0}^{n}\frac{(-1)^k}{\binom{n}{k}},n=1\right)=0+2+6+12+\ldots+2\binom{n}{2}=2\binom{n+1}{3}$$ so, by induction: $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}^{-1}=\sum_{h=0}^{n-1}\frac{(-1)^{h+1}(h+1)}{n-h}\binom{n+1}{h+2}=(n+1)\sum_{h=0}^{n-1}\frac{(-1)^{h+1}}{h+2}\binom{n}{h}$$ but the last sum is easily manageable through usual techniques, since it is clearly related with $\int_{0}^{1}x(1-x)^n\,dx$.

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  • $\begingroup$ What does it mean for a sum to have a complex number in the upper limit, as in $\sum_{k=0}^n\frac{(-1)^k}{\binom nk}$? Or is $n$ not the variable of the mentioned meromorphic function? $\endgroup$ Aug 19, 2015 at 19:09
  • $\begingroup$ @barto: to make it simpler, I am just considering the partial fraction decomposition of every term that appears in the original sum. $\endgroup$ Aug 19, 2015 at 19:23
  • $\begingroup$ @JackD'Aurizio What's "meromorphic function of $n$"? $\endgroup$
    – user196761
    Aug 20, 2015 at 14:27
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We seek to find a closed form of

$$\sum_{r=0}^n (-1)^r {n\choose r}^{-1}.$$

Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$

$$\frac{1}{k} {n\choose k}^{-1} = [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$

We get for our sum

$$1+ \sum_{r=0}^n (-1)^r r [z^n] \log\frac{1}{1-z} (z-1)^{n-r} \\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^n [v^n] \frac{1}{1-v} \sum_{r\ge 0} (-1)^r r (z-1)^{-r} v^r \\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^n [v^n] \frac{1}{v-1} \frac{v/(z-1)}{(1+v/(z-1))^2} \\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^{n+1} [v^{n-1}] \frac{1}{v-1} \frac{1}{(z-1+v)^2}.$$

The contribution from $v$ is

$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^n} \frac{1}{v-1} \frac{1}{(z-1+v)^2}.$$

The residue at infinity is zero so we may compute this by taking minus the residue at one and minus the residue at $v=1-z.$ (Residues sum to zero.) We get for the former

$$- [z^{n+2}] \log\frac{1}{1-z} (z-1)^{n+1} = - {n+2\choose 1}^{-1} = -\frac{1}{n+2}$$

and the latter

$$- \left.\left( \frac{1}{v^n} \frac{1}{v-1} \right)'\right|_{v=1-z} = - \left.\left( -\frac{n}{v^{n+1}} \frac{1}{v-1} - \frac{1}{v^n} \frac{1}{(v-1)^2} \right)\right|_{v=1-z} \\ = - \frac{n}{z (1-z)^{n+1}} + \frac{1}{(1-z)^n z^2}.$$

We get for the first term

$$- (-1)^{n+1} n [z^{n+1}] \log\frac{1}{1-z} = (-1)^n \frac{n}{n+1}.$$

and for the second

$$(-1)^n [z^{n+2}] \log\frac{1}{1-z} (z-1) = (-1)^n \frac{1}{n+1} - (-1)^n \frac{1}{n+2}.$$

Collecting everything we find

$$1+(-1)^n - \frac{1}{n+2} (1+(-1)^n).$$

This will produce zero when $n$ is odd. For $n$ even we find

$$\bbox[5px,border:2px solid #00A000]{ 2 \frac{n+1}{n+2}.}$$

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HINT:

In Brychkov, Marichev, Prudnikov Vol 1, 4.2.4 #2 there appears the more general sum:

$$\sum_{k=0}^n (-1)^k\binom{a}{k}^{-1}$$

We use the following formula

$$\binom{a}{k}^{-1} = \frac{a+1}{a+2}\left(\binom{a+1}{k}^{-1} + \binom{a+1}{k+1}^{-1}\right)$$

Now the alternating sum of terms from the LHS becomes a telescoping sum.

$\bf{Added:}$ The above quoted volume contains many other interesting results.

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