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The following problem is from a high school problem set. The students do not know how to integrate yet although they are comfortable with differentiation.

The problem is

Show that $\displaystyle S:=\sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k}}{n+k}= \frac{1}{n \binom{2n}{n}}$.

This problem can be solved by beta functions. It turns out that $$ S = \int_0^1 x^{n-1} (1-x)^n dx $$ when you expand the integrand using binomial theorem.

But are there any other, hopefully elementary, ways of solving the problem?

Isomorphism
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    https://math.stackexchange.com/questions/437523/proving-binomial-identity-without-calculus – lab bhattacharjee Apr 17 '21 at 01:50
  • I would like to note that the above linked problem is of a different nature than the one I have posted. The first problem in the linked post above was the previous problem in my problem set. I have solved that problem by elementary means and calculus. – Isomorphism Apr 17 '21 at 01:58
  • Dear Dave, I have already explained that it does not answer the question in my comment above. – Isomorphism Apr 18 '21 at 05:49

1 Answers1

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A common technique is to strengthen the induction hypothesis.

In particular, integrating and expanding $x^{m-1}(1-x)^n$ gives a hint that we can consider using: $$ \sum_{k=0}^n (-1)^k \frac{n \choose k}{m+k} = \frac{(m-1)!n!}{(m+n)!}$$

Let's do induction to show it's true for all $n$ starting at $0$.

If $n=0$, we have $\frac{1}{m}=\frac{(m-1)!0!}{m!}$.

Let $0 \leq n$. Let's show it being true for $n$ implies it's true for $n+1$.

We have starting with the LHS: $$\sum_{k=0}^{n+1} (-1)^k \frac{n+1 \choose k}{m+k}=\sum_{k=0}^{n+1} (-1)^k \frac{{n \choose k}+{n\choose k-1}}{m+k}$$ $$=\sum_{k=0}^{n} (-1)^k \frac{{n \choose k}}{m+k}-\sum_{k=0}^{n} (-1)^k \frac{{n\choose k}}{m+1+k} $$

Using the inductive hypothesis for $n$: $$=\frac{(m-1)!n!}{(m+n)!}-\frac{m!n!}{(m+n+1)!}$$ $$=\frac{(m-1)!n!}{(m+n)!} (1 -\frac{m}{m+n+1})$$ $$=\frac{(m-1)!n!}{(m+n)!} \frac{n+1}{m+n+1}$$ $$=\frac{(m-1)!(n+1)!}{(m+n+1)!}$$ Thus, if it's true for a given $n$, then it's true for $n+1$, finishing the inductive proof.

In particular, it's true when $m=n-1$ proving your initial equality.

Eric
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