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On p. 146 of Principles of Mathematical Analysis, Rudin claims that $\int_{0}^{1} x(1-x^2)^ndx=\frac{1}{2n+2}$. But evaluating the integral I found it equal to $$\int_{0}^{1}x(\sum\limits_{k=0}^{n}{{n}\choose{k}}(-x^2)^k)dx=\int_{0}^{1}\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k+1}dx=\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}\frac{x^{2k+2}}{2k+2}$$ evaluated at $x=1$, which equals $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$. So now I'm just wondering, why is this equal to $\frac{1}{2n+2}$?

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    If you substitute $x^2\mapsto u$ in the integral, then I'd say you've just proven the identity in your question. – 2018-03-26

4 Answers 4

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$$\begin{align*} \sum_{k=0}^n\binom{n}k\frac{(-1)^k}{2k+2}&=\frac{1}2\sum_{k=0}^n\frac{1}{k+1}\binom{n}k(-1)^k\\ &=\frac{1}2\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}(-1)^k\\ &=\frac{1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^{k-1}\\ &=\frac{-1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}\\ &=\frac{-1}{2n+2}\left(\sum_{k=0}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}-\binom{n+1}0(-1)^01^{n+1}\right)\\ &=\frac{-1}{2n+2}\left((-1+1)^{n+1}-1\right)\\ &=\frac{1}{2n+2} \end{align*}$$

by the binomial theorem.

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    @Dr.MV: No, actually, since I started writing it before you posted and didn't see your hint until after I finished writing, editing, and posting the final version. – 2017-01-10
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    No worry. You're the most prolific user herein. I was merely pointing out the flow of presentation is aligned with the hint I posted. And Happy New Year! -Mark – 2017-01-10
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    @Dr.MV: Same to you! (One-finger typing on a Kindle tablet is painfully slow, especially when you're typing mathematics.) – 2017-01-10
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    I completely understand. I'm on a "smart phone" and watching the championship football game. – 2017-01-10
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HINT:

Note that we can write

$$\frac{1}{2k+2}\,\binom{n}{k}=\frac{1}{2n+2}\binom{n+1}{k+1}$$

Then, substitute this into the summation, shift the index from $k$ to $k'=k+1$ so the new summation limits are $k'=1$ to $k'=n+1$, and use $\sum\limits_{k=0}^{n+1} \binom{n}{k}(-1)^k=0$.

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    There must be a typo. Did you mean $\binom{n}{k}=\frac{2k+2}{2n+2}\binom{n+1}{k+1}$ or $\frac1{2k+2}\binom{n}{k}=\frac1{2n+2}\binom{n+1}{k+1}$? – 2017-01-10
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    @robjohn Indeed. I meant the latter. Thank you for catching this; I've edited. And Happy New Year. -Mark – 2017-01-10
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    Hmm, following your hint we have $\sum_{k=0}^{n}{{n}\choose{k}}\frac{(-1)^k}{2k+2}=\sum_{k'=1}^{n+1}{{n+1}\choose{k+1}}\frac{(-1)^k}{2n+2}$ (noting that k'=k+1) $=\frac{1}{2n+2}\sum_{k'=1}^{n+1}{{n+1}\choose{k'}}(-1)^{k'-1}=\frac{1}{2n+2}\sum_{k=0}^{n}{{n+1}\choose{k+1}}(-1)^{k}$. Now I'm not sure exactly where $\sum_{k=0}^{n+1}{{n}\choose{k}}(-1)^k=0$ comes in, or why this identity holds. We have $\sum_{k=0}^{n+1}{{n}\choose{k}}(-1)^k=1-n+{{n}\choose{2}}- \dots + {{n}\choose{n+1}}$ (assuming n+1 even). Why does this equal 0 (and if it does, how does that help us)? – 2017-01-10
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    @Rasputin Note that $$\sum_{k'=1}^{n+1}\binom{n+1}{k'}(-1)^{k'-1}=1+\sum_{k'=0}^{n+1}\binom{n+1}{k'}(-1)^{k'-1}=1$$The sum from $k'=0$ since it is the binomial expansion of $(-1)\times (1+(-1))^{n+1}$. – 2017-01-10
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    Great, thank you! That makes perfect sense. Guess I will just keep trying to prove $\sum_{k=0}^{n+1}{{n+1}\choose{k}}(-1)^k=0$ then . . . – 2017-01-10
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    You're welcome! My pleasure. I would suggest using induction to prove the binomial theorem. And wishing you a very Happy New Year! -Mark – 2017-01-10
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    Right, got it! Thanks, to you as well! – 2017-01-10
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Why not use the substitution $1-x^{2}=t$ to change the integral to $$\frac{1}{2}\int_{0}^{1}t^{n}\,dt=\frac{1}{2n+2}$$ But you seem more interested in the link between your sum and value of integral in a direct manner as given in answers by Brian Scott and Dr. MV.

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    Thanks! I'm very interested in more simple solutions as well. Just wondering though, how does this substitution change the integral to $\frac{1}{2}\int_{0}^{1}t^ndt$? Wouldn't it change the integral to $\int_{0}^{1}(\sqrt{1-t})(t^n)dt$ (or something like that -- I'm not too sure how to deal with whether it's $+ \sqrt{1-t}$ or $-\sqrt{1-t}$ . . .)? – 2017-01-10
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    @Rasputin: differentiate $1-x^{2}=t$ to get $-2x\,dx=dt$ and hence the part $x\, dx$ gets replaced by $-dt/2$. There is no need to replace just $x$ by $\sqrt{1-t}$. – 2017-01-10
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    Ah right, that makes sense, thank you! Would we then have that it equals $-\frac{1}{2n+2}$ though because $x$ gets replaced by $-dt/2$ instead of just $dt/2$? – 2017-01-10
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    @Rasputin : note that the limits $0,1$ for $x$ correspond to limits $1,0$ for $t$ and reversing them again to $0,1$ adds one more minus sign which fixes $-dt/2$. – 2017-01-10
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    Ah right that makes sense, thank you! – 2017-01-10
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An overkill. The Melzak's identity tell us that $$f\left(x+y\right)=y\dbinom{y+n}{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{f\left(x-k\right)}{y+k} $$ where $f$ is an algebraic polynomial up to dergee $n$, x,y\in\mathbb{R},\, y\neq-1,\dots-n . So taking $f\equiv1 $ and $y=1$ we get $$1=\dbinom{1+n}{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{1+k} $$ hence $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{1+k}=\color{red}{\frac{1}{n+1}}\tag{1} $$ now you have only to multiply both side of $(1)$ by $1/2$.