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.

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

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

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

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

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

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

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

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.

Ishan Singh - 3 years, 12 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.

Kartik Sharma - 3 years, 12 months ago

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

Ishan Singh - 3 years, 12 months ago

@Ishan Singh – 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.

Kartik Sharma - 3 years, 12 months ago

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

Ishan Singh - 3 years, 12 months ago

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

Kartik Sharma - 3 years, 12 months ago

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

Ishan Singh - 3 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$