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# Binomial Coefficient Challenge 4!

Find a closed form of

$\displaystyle \sum_{k=0}^{n-2} {\binom{k}{m-1} \frac{1}{n-k-1}}$

Note by Kartik Sharma
2 months, 2 weeks ago

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For $$n<m$$, the sum is $$0$$. Also, for $$m=1$$, the sum is $$H_{n-1}$$. Henceforth, we'll assume $$n \geq m>1$$.

Denote the sum as $$A_{n}$$. Note that,

$\dfrac{n-1}{k} \sum_{j=1}^{k} \dfrac{1}{(j-n)(j-n+1)} = \dfrac{1}{n-k-1}$

We have,

$A_{n} = \sum_{k=0}^{n-2} {\dbinom{k}{m-1} \dfrac{1}{n-k-1}}$

$= (n-1) \sum_{k=1}^{n-2} \sum_{j=1}^{k} \dfrac{1}{k} \dbinom{k}{m-1} \left( \dfrac{1}{(j-n)(j-n+1)} \right)$

$= \left(\dfrac{n-1}{m-1}\right) \sum_{j=1}^{n-2} \sum_{k=j}^{n-2} \dfrac{1}{(j-n)(j-n+1)} \dbinom{k-1}{m-2}$

$= \left(\dfrac{n-1}{m-1}\right) \sum_{j=1}^{n-2} \left( \dfrac{1}{(j-n)(j-n+1)} \sum_{k=j}^{n-2} \dbinom{k-1}{m-2} \right)$

$= \left(\dfrac{n-1}{m-1}\right) \sum_{j = 1}^{n-2} \dfrac{1}{(j-n)(j-n+1)} \left( \dbinom{n-2}{m-1} - \dbinom{j-1}{m-1} \right)$

$= \text{X} + \text{Y}$

Now,

$\text{X} = \left(\dfrac{n-1}{m-1}\right) \dbinom{n-2}{m-1} \sum_{j=1}^{n-2} \dfrac{1}{(j-n)(j-n+1)}$

$= \left( \dfrac{n-2}{m-1} \right) \dbinom{n-2}{m-1}$

Also,

$\text{Y} = \left(\dfrac{n-1}{m-1}\right) \left[ \sum_{j=1}^{n-2} \dfrac{1}{n-j} \dbinom{j-1}{m-1} - \sum_{j=1}^{n-2} \dfrac{1}{n-j-1} \dbinom{j-1}{m-1} \right]$

Re-indexing, we have,

$\text{Y} = \left(\dfrac{n-1}{m-1}\right) \left(A_{n} - A_{n-1} - \dbinom{n-2}{m-1} \right)$

So that,

$A_{n} = \text{X} + \text{Y}$

$\implies A_{n} = \left(\dfrac{n-1}{m-1}\right) [A_{n} - A_{n-1}] - \dfrac{1}{m-1} \dbinom{n-2}{m-1}$

$\implies \dfrac{(n-m)!}{(n-1)!} A_{n} - \dfrac{(n-m-1)!}{(n-2)!} A_{n-1} = \dfrac{(n-m-1)!}{(n-1)!} \dbinom{n-2}{m-1}$

$\implies \sum_{n=m}^{N} \left(\dfrac{(n-m)!}{(n-1)!} A_{n} - \dfrac{(n-m-1)!}{(n-2)!} A_{n-1} \right) = \sum_{n=m}^{N} \dfrac{(n-m-1)!}{(n-1)!} \dbinom{n-2}{m-1}$

$\implies \dfrac{(N-m)!}{(N-1)!} A_{N} = \dfrac{1}{(m-1)!} \sum_{k=m}^{N} \dfrac{1}{k}$

$\implies A_{N} = \dbinom{N-1}{m-1} (H_{N-1} - H_{m-1})$

$\therefore \sum_{k=0}^{n-1} \dbinom{k}{m} \dfrac{1}{n-k} = \dbinom{n}{m} (H_{n} - H_{m}) \quad \square$ · 2 months ago

@Kartik Sharma What was your approach? · 2 months ago

Same as yours. Just a little different formatting. · 2 months ago

Very elegant problem. I'm getting the closed form as $\dbinom{n-1}{m-1} (H_{n-1} - H_{m-1})$

Reformulating it, we get the elegant identity :

$\sum_{k=0}^{n-1} \dbinom{k}{m} \dfrac{1}{n-k} = \dbinom{n}{m} (H_{n} - H_{m})$

For $$m,n \in \mathbb{Z^{+}}$$

Will post full solution soon. · 2 months, 1 week ago

Correct! It's very elegant indeed. · 2 months, 1 week ago

Can u tell me any name of book or any resource on internet to learn about more binomial co-efficient ??/ · 2 months, 1 week ago

Sure. I have mostly learnt from Concrete Mathematics. Apart from that, binomial coefficients are a good application of algebra, calculus, recurrence etc. · 2 months, 1 week ago

Thanks for the reference · 2 months, 1 week ago