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so I am supposed to solve a proof which seems fairly easy, but the negative exponents in $$\sum_{k=0}^n \binom nk\ (\frac{(-1)^k}{k+1})= \frac{1}{n+1}$$ are making this question very difficult for me. I have tried using binomial theorem on the right side with $(n+1)^{-1}$ but I understand that operation does not make sense. I can also tell that the only difference between the two proofs is that the left side has an additional $(k+1)^{-1}$ and the right side has an additional $(n+1)^{-1}$, but I am still having difficulty solving this question. Hints appreciated.

discombobulator
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2 Answers2

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Hint: Consider antidifferentiating $f(x)=(1-x)^n$.

Arthur
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  • I am not at the level where I have learned integration. – discombobulator May 30 '19 at 11:24
  • No integration here. Just differentiating in reverse. – Arthur May 30 '19 at 11:24
  • I have not learned antidifferentiation as well. The only knowledge I have of calculus barely reaches differentiation. – discombobulator May 30 '19 at 11:25
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    Well, then this approach is going to be a bit more difficult. – Arthur May 30 '19 at 11:31
  • @Arthur I actually don't see what to do. You get $\sum_{k=0}^n {n \choose k}(-1)^k \frac{x^{k+1}}{k+1} = -\frac{1}{n+1}(1-x)^{n+1}$. Now what value of $x$ do you plug in? – mathworker21 May 30 '19 at 11:35
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    @matgworker21 You plug in $x=1$, since that makes the terms in the sum what we want them to be. However, you've committed one of the classic mistakes: You didn't pay enough attention to the constant term of the antiderivative. – Arthur May 30 '19 at 12:02
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The binomial theorem gives: $$(1-x)^n= \sum_{k=0}^{n} (-1)^k x^k$$ Integrate w.r.t. x both sides from $x=0$ to $x=1$, to get $$\left .\sum_{k=0}^{n} (-1)^k {n \choose k}\frac{x^{k+1}}{k+1} \right|_{0}^{1}= -\left.\frac{(1-x)^{n+1}}{n+1} \right|_{0}^{1}=\frac{1}{n+1}.$$

Z Ahmed
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  • I don't know if you noticed my answer and the comments below it, but this doesn't seem to be an approach that the OP is after, or even understands. – Arthur May 30 '19 at 12:16