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.*

## Comments

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TopNewest\[\text{S}=\displaystyle\sum_{r=0}^{m-1}\dfrac{(-1)^r}{\dbinom{n}{r}}\]

\[=\displaystyle\sum_{r=0}^{m-1}(-1)^r\dfrac{r!(n-r)!}{n!}\]

\[=\displaystyle\sum_{r=0}^{m-1}(-1)^r \dfrac{r!(n-r)!\color{blue}{(n-r+1+r+1)}}{n!}\color{red}{\times\dfrac{1}{(n+2)}}\]

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

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

where \(T_{r} = (-1)^r r!(n-r+1)! \)

Evaluating using telescoping the sum, we have,

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

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

\[\dfrac{1}{\binom{n}{k}}=\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}\dfrac{j}{n-j+1}\]

The continued fraction result is fascinating! I'll work on it when I get some time. – Ishan Singh · 2 months, 1 week ago

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@Kartik Sharma Btw, what was your approach to this problem? – Ishan Singh · 2 months, 1 week ago

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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 · 2 months, 1 week ago

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Gosper's method?

I was talking about the summation. I assume thisI have added some explanation in the telescoping part for clarity – Ishan Singh · 2 months, 1 week ago

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– Kartik Sharma · 2 months, 1 week ago

Yes. It is quite general.Log in to reply