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I'm wondering if there is an natural way to see that

$\sum_0^n {n\choose k}(-1)^k \frac{1}{2k + 2} = 1/(2n + 2)$ for $n\gt 0$.

Both sides of the equation are equal to a simple integral Integral of $x(1-x^2)^n$

By "natural" I mean that you would be able to deduce the right hand side without knowing it previously. For example, an induction argument is slightly unnatural since we would have to guess the closed form beforehand.

Community
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Mark
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  • It looks like the identity involves noticing that alternating terms cancel in such a way as to produce the result. Have you tried canceling terms in this way? – abiessu Jan 01 '15 at 03:15
  • @Mark, Sorry, I must have read your post carelessly... – Sangchul Lee Jan 01 '15 at 03:43

3 Answers3

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$$\binom nk\frac1{2(k+1)}=\frac12\cdot\frac{n!}{(n-k)!\cdot k!\cdot(k+1)}$$

$$=\frac1{2(n+1)}\cdot\frac{(n+1)!}{[n+1-(k+1)]!\cdot(k+1)!}=\frac1{2n+2}\binom{n+1}{k+1}$$

So, $\displaystyle\sum_{k=0}^n\binom nk(-1)^k\frac1{2k+2}=\frac1{2n+2}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}$

If $\displaystyle S=\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=-\sum_{k=0}^n(-1)^{k+1}\binom{n+1}{k+1}$

Setting $k+1=r,$

$$S=-\sum_{r=1}^{n+1}(-1)^r\binom{n+1}r\iff-S=\sum_{r=1}^{n+1}(-1)^r\binom{n+1}r$$

$$-S=\sum_{r=0}^{n+1}(-1)^r\binom{n+1}r-\binom{n+1}0(-1)^0=(1-1)^{n+1}-1$$

lab bhattacharjee
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  • This is of course a much better way than attempting to cancel alternating terms, but for the record I have applied alternating-term-cancelation and it works too. :-) – abiessu Jan 01 '15 at 03:23
  • @abiessu, Would you mind sharing you method? – lab bhattacharjee Jan 01 '15 at 03:26
  • I wouldn't mind, but it's very messy. I'll see what I can do. – abiessu Jan 01 '15 at 03:29
  • This is very nice. I suspected that the identity $(1-1)^N = \sum {N\choose k}(-1)^k$ would appear somewhere but unfortunately I tried to get to it by summing by parts. – Mark Jan 01 '15 at 03:34
  • @Mark It's the binomial theorem $(x + y)^n$ where $y = -y_0$. Every term with odd power of $y$ is negative because $y$ is negative. So, when we let $x = y_0 = 1$, all we're left with is the binomial coefficients. However, at each odd power of $y$, the coefficient is negative. – Axoren Jan 01 '15 at 03:40
  • @Axoren you mean $y = -x$ right? Yes I was saying I was already aware of this equality and I tried to force it into the equation somewhere by using summation by parts. – Mark Jan 01 '15 at 03:42
  • @Mark My apologies; I misunderstood your comment. – Axoren Jan 01 '15 at 03:44
  • @Axoren no worries hopefully your comment may be useful for someone else. – Mark Jan 01 '15 at 03:45
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$$(1+x)^n=\sum_{r=0}^n\binom nrx^r$$

Integrate both sides to get $$\frac{(1+x)^{n+1}}{n+1} =\sum_{r=0}^n\binom nr\frac{x^{r+1}}{r+1}+K $$ where $K$ is an arbitrary constant

Set $x=0$ in the above identity to determine $K$

Finally set $x=-1$

lab bhattacharjee
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  • This is very slick. Is this an application of some more general technique or is it ad hoc for this problem? – Mark Jan 01 '15 at 05:50
  • @Mark, See http://math.stackexchange.com/questions/437523/proving-binomial-idenity-without-calculus – lab bhattacharjee Jan 01 '15 at 06:04
  • I think this is similar to your other solution. In your link I don't see where the integral comes in to play. In fact the question asks for solutions to avoid calculus? – Mark Jan 01 '15 at 06:11
  • @Mark, It was posted by me:) My point was : these types of Questions are ideally handled by calculus. – lab bhattacharjee Jan 01 '15 at 06:13
  • Oh I see! Okay I will try to figure out the calculus solutions to your question – Mark Jan 01 '15 at 18:10
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You want to prove

$$\sum_{k=0}^n {n\choose k} \frac{1}{k + 1} (-1)^k= \frac{1}{n + 1}$$

i.e.

$$\sum_{k=0}^n {n\choose k}\frac{n+1}{k + 1} (-1)^k = 1$$

which is

$$\sum_{k=0}^n {n+1\choose k+1}(-1)^k = 1$$

which is

$$\frac{1}{X}\sum_{k=0}^{n+1} {n+1\choose k+1}X^{k+1}\left.\right|_{X=\,-1} = 1$$

As $\ \ (X+Y)^n=\sum_{k=0}^n\binom nk X^{n-k}Y^k\ \ $ we find the left hand side is

$$\frac{1}{X}\left((X+1)^{n+1}-1\right)\left.\right|_{X=\,-1}$$

and that's $1$ indeed.

Nikolaj-K
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  • Nice except that you kind of "guessed" the correct form of the solution when you started out. – Mark Jan 01 '15 at 05:47
  • @Mark: Yeah, I interpreted OPs question "I'm wondering if there is an natural way to see that $a=b$" not as task to simplify $a$, but to interpret the equation. – Nikolaj-K Jan 01 '15 at 12:55