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I have been given a problem as part of a practice set for a problem solving class. Here, I am asked to prove the following equality (for positive integers $m$ and $n$):

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

I have tried to form a counting argument,though that doesn't seem to be working too well. The theme of these problems is that they can be re-written as simpler problems, although I am just not sure how to do that here.

Some other things I tried include rewriting each sum as the difference of two other sums (ie putting the positive elements into one sum and the negatives into another) but that doesn't seem to simplify it at all...

Any help is much appreciated

myriagon
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  • The sum on the LHS of the identity to prove is $$\sum_{k=0}^n(-1)^k{n\choose k}\int_0^1x^{k+m}dx=\int_0^1x^m\sum_{k=0}^n(-1)^k{n\choose k}x^kdx=\int_0^1x^m(1-x)^ndx$$ Can you show that the integral on the RHS above is symmetric in $(n,m)$? – Did Jan 22 '18 at 23:06
  • Both sides of your equality are equal to $\dfrac{1}{\left(n+m+1\right) \dbinom{n+m}{m}}$; this follows easily from https://math.stackexchange.com/questions/2455428/an-identity-involving-binomial-coefficients-and-rational-functions . – darij grinberg Jan 22 '18 at 23:19

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A nice technique for this kind of thing is to involve integrals: $$\eqalign{\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{k+m+1} &=\sum_{k=0}^{n}(-1)^k\binom{n}{k}\int_0^1x^{k+m}\,dx\cr &=\int_0^1\sum_{k=0}^{n}(-1)^k\binom{n}{k}x^{k+m}\,dx\cr &=\int_0^1 x^m(1-x)^n\,dx\cr &=\int_0^1 y^n(1-y)^m\,dy\qquad\hbox{(substitute $y=1-x$)}\cr &=\cdots\qquad\hbox{(similar calculation, I'll leave it to you)}\cr &=\sum_{k=0}^{m}(-1)^k\binom{m}{k}\frac{1}{k+n+1}\ .\cr}$$

David
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  • I get mostly all of this, so thank you. But I have a quick question:

    When you substitute in $y=1-x$ , shouldnt the limits of integration be 1 to 0 not 0 to 1? Why do they stay the same?

     – myriagon Jan 23 '18 at 03:07
  • Because $dy=-dx$ which swaps the limits back again. – David Jan 23 '18 at 03:17
  • Wow...silly oversight by me. Thanks a lot for your help. – myriagon Jan 23 '18 at 03:20