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I need to prove the following: $$\sum_{k=0}^n {n \choose k}\cdot\frac{(-1)^k}{k+1} = \frac{1}{n+1}$$

Started from the left, I got this far: $$\frac{1}{n+1}+\sum_{k=1}^n {n+1 \choose k+1}\cdot(-1)^k$$

Am I right so far? Any ideas how to move on? thanks.

Edit: I guess my last move had a mistake, then perhaps this is ok? $$\frac{1}{n+1}\cdot\sum_{k=0}^n {n+1 \choose k+1}\cdot(-1)^k$$

How do you suggest I move on?

Martin Sleziak
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user401516
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2 Answers2

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Integrate both sides of $$\sum_{k=0}^n{n\choose k}x^k=(x+1)^n$$ to obtain that $$C+\sum_{k=0}^n\frac{1}{k+1}{n\choose k}x^{k+1}=\frac{1}{n+1}(x+1)^{n+1}$$ Where $C$ is a constant of integration which equals $\frac{1}{n+1}$ upon substituting $x=0$. For your identity, substitute $x=-1$.

ArtW
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Note that we have $$ \binom{n}{k}\frac{(-1)^k}{k+1} = \frac{n!}{k!(n-k)!}\frac{(-1)^k}{k+1} = (-1)^k\frac{n!}{(k+1)!(n-k)!} =\\ \frac{(-1)^k}{n+1}\frac{(n+1)!}{(k+1)!(n-k)!} = \binom{n+1}{k+1}\frac{(-1)^k}{n+1}\\ = \frac{-1}{n+1}\cdot (-1)^{k+1}\binom{n+1}{k+1} $$ Now, while keeping the indices straight, compare the sum of these terms to what the binomial theorem says about $$ \frac{(1-1)^{n+1}}{n+1} $$

Edit: More details.

The binomial theorem says that $$ 0 = \frac{(1-1)^{n+1}}{n+1} = \sum_{m = 0}^{n+1}\frac{1}{n+1}\binom{n+1}{m}\tag1(-1)^{m} $$ while after the transformation at the top of my answer, your sum looks like $$ \sum_{k = 0}^n\frac{-1}{n+1}\binom{n+1}{k+1}(-1)^{k+1} $$ Doing the index transformation $m = k+1$, this last one becomes $$ \sum_{m = 1}^{n+1}\frac{-1}{n+1}\binom{n+1}{m}(-1)^{m}\tag2 $$ Now can you compare the sums $(1)$ and $(2)$?

Arthur
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