Binomial Coefficient Challenge 6!

Prove the following -

\displaystyle \begin{equation*} 1 - \cfrac{1}{n+\cfrac{2n}{n-3+\cfrac{3(n-1)}{n-5+\cfrac{4(n-2)}{\ddots \cfrac{(m-1)(n-m+3)}{n-(2m-3)}}}}} \end{equation*} = \sum_{0\le k <m}{\dfrac{{(-1)}^k}{\binom{n}{k}}} = \dfrac{n+1}{n+2}\left(1 + \frac{{(-1)}^{m+1}}{\binom{n+1}{m}}\right)

The continued fraction part is optional. It is just a bonus.

This is completely original.

Note by Kartik Sharma
3 years, 4 months ago

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1 vote

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S=r=0m1(1)r(nr)\text{S}=\displaystyle\sum_{r=0}^{m-1}\dfrac{(-1)^r}{\dbinom{n}{r}}

=r=0m1(1)rr!(nr)!n!=\displaystyle\sum_{r=0}^{m-1}(-1)^r\dfrac{r!(n-r)!}{n!}

=r=0m1(1)rr!(nr)!(nr+1+r+1)n!×1(n+2)=\displaystyle\sum_{r=0}^{m-1}(-1)^r \dfrac{r!(n-r)!\color{#3D99F6}{(n-r+1+r+1)}}{n!}\color{#D61F06}{\times\dfrac{1}{(n+2)}}

=r=0m11n!(n+2)[(1)r{r!(nr+1)!+(nr)!(r+1)!}]=\displaystyle\sum_{r=0}^{m-1}\dfrac{1}{n!(n+2)}\left[(-1)^r\left\{r!(n-r+1)!+(n-r)!(r+1)!\right\}\right]

=r=0m11n!(n+2)[TrTr+1]=\displaystyle\sum_{r=0}^{m-1}\dfrac{1}{n!(n+2)}\left[T_{r} - T_{r+1}\right]

where Tr=(1)rr!(nr+1)!T_{r} = (-1)^r r!(n-r+1)!

Evaluating using telescoping the sum, we have,

S=n+1n+2(1+(1)m+1(n+1m))\text{S} = \dfrac{n+1}{n+2}\left(1+\dfrac{(-1)^{m+1}}{\dbinom{n+1}{m}}\right)

Some other methods can be to use Beta functions or using the result (from Partial Fractions)

1(nk)=j=1k(1)kj(kj)jnj+1\dfrac{1}{\dbinom{n}{k}}=\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}\dfrac{j}{n-j+1}

The continued fraction result is fascinating! I'll work on it when I get some time.

Ishan Singh - 3 years, 4 months ago

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@Kartik Sharma Btw, what was your approach to this problem?

Ishan Singh - 3 years, 4 months ago

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I used Finite Calculus(Gosper's Method). I thought of beta functions but it was tedious. Telescoping serious was very nice although a bit too tricky.

I will add my method later. For the continued fraction part if you are asking, then I used a standard result in the theory of evaluating continued fractions.

Kartik Sharma - 3 years, 4 months ago

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@Kartik Sharma I was talking about the summation. I assume this Gosper's method?

I have added some explanation in the telescoping part for clarity

Ishan Singh - 3 years, 4 months ago

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@Ishan Singh Yes. It is quite general.

Kartik Sharma - 3 years, 4 months ago

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