# Binomial Coefficient Challenge 9!

Find a closed form of

$\displaystyle \sum_{k=0}^{n-1}{\binom{k}{m} H_k}$

where $$H_k$$ is the $$k$$th Harmonic number.

Note by Kartik Sharma
1 year, 3 months ago

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$\sum_{k=0}^{n} \dbinom{k}{m} H_{k} = \sum_{k=1}^{n} \dbinom{k}{m} H_{k}$

$= \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{r} \dbinom{k}{m}$

$= \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r} \dbinom{k}{m}$

$= \sum_{r=1}^{n} \dfrac{1}{r} \left(\dbinom{n+1}{m+1} - \dbinom{r}{m+1}\right)$

$= \dbinom{n+1}{m+1} H_{n} - \dfrac{1}{m+1} \sum_{r=1}^{n}\dbinom{r-1}{m}$

$= \dbinom{n+1}{m+1} H_{n} - \dfrac{1}{m+1} \dbinom{n}{m+1}$

$\therefore \sum_{k=0}^{n} \dbinom{k}{m} H_{k} = \dbinom{n+1}{m+1}\left(H_{n+1} - \frac{1}{m+1}\right) \ \square$

Alternatively, we can form a recurence in $$n$$ or $$m$$ and solve for the closed form. Yet another approach can be to consider the identity $\sum_{r=0}^{k-1} \dbinom{r}{m-1} = \dbinom{k}{m}$ and substitute in the sum.

Note that this result also holds for values of $$m$$ other than positive integers.

- 1 year, 3 months ago

There is a typo. It should be $$\binom{n}{m+1}$$ rather than $$\binom{n+1}{m+1}$$. The sum is till $$n-1$$.

Easiest of the lot, I guess. Yet another method is to use summation by parts. It was intended for that.

- 1 year, 3 months ago

Beside $$H_{n}$$? I have checked numerically, seems fine. Btw, my method is Summation By Parts only, if you look closely.

- 1 year, 3 months ago

The question is $$\displaystyle \sum_{k=0}^{n-1}{\binom{k}{m} H_k}$$ and not $$\displaystyle \sum_{k=0}^{n}{\binom{k}{m} H_k}$$.

Oh, indeed now I see. I was looking for the formal method, lol.

- 1 year, 3 months ago

Yes. But both sums imply each other :) Anyway, waiting for the next part.

- 1 year, 3 months ago

Then the answer is $$\displaystyle \binom{n}{m+1}\left(H_n - \frac{1}{m+1}\right)$$ which looks nice.

- 1 year, 3 months ago

Yeah, it can further be tidied up. I'll edit it. Btw, check out my solution of the latest problem based on Challenge 4.

- 1 year, 3 months ago