Summation proof

[Albert Chan]

(09-11-2023 03:48 PM)Albert Chan Wrote:  \(\displaystyle \binom{x+n}{k} =
\left(\frac{x+n}{k}\right) \binom{x+n-1}{k-1}
\)

\(\displaystyle \binom{x+n}{k}' \bigg\rvert_{x=0} =
\left(\frac{1}{k}\right) \binom{n-1}{k-1} +
\left(\frac{n}{k}\right) \binom{x+n-1}{k-1}'\bigg\rvert_{x=0}
\)

I had managed to undo recursion, into iterative sums (written prove welcome!)

\(\displaystyle \binom{x+n}{k}' \bigg\rvert_{x=0}
\;=\; \sum_{j=1}^{k} \frac{(-1)^{j+1}}{j}\; \binom{n}{k-j}
\;=\; \binom{n}{k}\;\sum_{j=n-(k-1)}^{n} \frac{1}{j}
\)

I was typing the question to Math StackExchange, then I figured out the proof
It is a bit OT to Bernoulli's number, so I start a new thread, for those interested.

Proving RHS is easy, if we noticed \(\displaystyle \left(\frac{1}{k}\right) \binom{n-1}{k-1} = \left(\frac{1}{n}\right) \binom{n}{k}\)

Also, if k is negative, comb(n, k) = 0. This explained why summation has exactly k terms.

\(\begin{align}
\displaystyle \binom{x+n}{k}' \bigg\rvert_{x=0} &=
\frac{\binom{n}{k}}{n} +
\left(\frac{n}{k}\right) \left(
\frac{\binom{n-1}{k-1}}{n-1} + \left(\frac{n-1}{k-1}\right)
\left(\frac{\binom{n-2}{k-2}}{n-2} + \left(\frac{n-2}{k-2}\right) (...) \right)
\right) \\
&= \binom{n}{k} \left(\frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2}+\;...\;+
\frac{1}{n-(k-1)} \right) \\
&= \binom{n}{k}\;\sum_{j=n-(k-1)}^{n} \frac{1}{j}
\end{align}\)