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$$\sum _{k=0}^n (-1)^k \frac{1}{\binom n k}=\frac{n+1}{n+2} (1+(-1)^n)$$

$$A(n,k)=(-1)^k {\binom n k}^{-1}=(-1)^k \frac{(n-k)!k!}{n!}$$

$$A(n+1,k+1)-A(n+1,k)=-\frac{n+2}{n+1} A (n,k)$$

$$\sum_{k=1}^n A(n,k)=-\frac{n+1}{n+2} (A(n+1,n+1)-A(n+1,0)) = (1+(-1)^n) \: \frac{n+1}{n+2}$$

Is my approach correct?

Did
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    If you want to see $\displaystyle \left( \frac 1 2 \right)$ rather than $\displaystyle ( \frac 1 2 )$ you need to write \left( \frac 1 2 \right) rather than ( \frac 1 2 ). But in things like $A(n,k)$ and $(-1)^n$, the use of \left and \right seems like just clutter that makes editing more onerous. Also, note the use of \binom, as in my edits. You don't need that complicated stuff you did instead. $\qquad$ – Michael Hardy Nov 23 '16 at 19:51
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    Try induction perhaps? – Simply Beautiful Art Nov 23 '16 at 20:04

2 Answers2

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Another approach $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}} &=(n+1)\sum_{k=0}^n(-1)^k\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}\tag{1}\\ &=(n+1)\sum_{k=0}^n(-1)^k\int_0^1t^k(1-t)^{n-k}\,\mathrm{d}t\tag{2}\\ &=(n+1)\int_0^1\frac{1-(-1)^{n+1}\frac{t^{n+1}}{(1-t)^{n+1}}}{1+\frac{t}{1-t}}(1-t)^n\,\mathrm{d}t\tag{3}\\ &=(n+1)\int_0^1\left((1-t)^{n+1}-(-1)^{n+1}t^{n+1}\right)\,\mathrm{d}t\tag{4}\\[4pt] &=\frac{n+1}{n+2}\left(1-(-1)^{n+1}\right)\tag{5} \end{align} $$ Explanation:
$(1)$: write the binomial coefficients as ratios of Gamma Functions
$(2)$: Use the Beta Function integral
$(3)$: apply the Formula for the Sum of a Geometric Series
$(4)$: simplify
$(5)$: integrate

robjohn
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Your approach is nice and correct. It's a good idea to simplify the sum with the technique of telescoping.

Two aspects:

  • Since your proof is valid for all non-negative integers $n$ you should clearly state the range of validity.

  • Since the approach of telescoping is valid and sufficient to prove the claim no proof by induction is necessary.

Typo: The sum in the last line should start with $k=0$.

Community
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Markus Scheuer
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