Thread: Prove that series = ... ?

1. Prove that series = ... ?

Prove that:

$\displaystyle 1 - \frac{n^2}{n + 1} + \frac{n^2(n^2 - 1)}{(n + 1)(n + 2)} - \frac{n^2(n^2 - 1)(n^2 - 2)}{(n + 1)(n + 2)(n + 3)} + \mbox{...}\ = \frac{1}{n + 1}$

where $\displaystyle n\in \mathbb{N}$

2. Note that you may re-write the sum as: $\displaystyle S_n=\sum_{k=0}^{\infty}{\tfrac{\binom{n^2}{k}}{\bi nom{n+k}{k}}\cdot (-1)^k}$

Now: $\displaystyle n \cdot \int_0^1 {x^k \cdot \left( {1 - x} \right)^{n - 1} dx} =\tfrac{1}{\binom{n+k}{k}}$ (you should recognise the Beta function here)

So:$\displaystyle \tfrac{1} {n} \cdot S_n = \sum\limits_{k = 0}^\infty {\binom{n^2}{k} \cdot \left( {\int_0^1 {x^k \cdot \left( {1 - x} \right)^{n - 1} dx} } \right) \cdot \left( { - 1} \right)^k }$$\displaystyle = \int_0^1 {\left( {\sum\limits_{k = 0}^\infty {\binom{n^2}{k} \cdot x^k \cdot \left( { - 1} \right)^k } } \right) \cdot \left( {1 - x} \right)^{n - 1} dx}$

Now by the Binomial Theorem: $\displaystyle \tfrac{1} {n} \cdot S_n = \int_0^1 {\left( {1 - x} \right)^{n^2 } \cdot \left( {1 - x} \right)^{n - 1} dx} = \tfrac{1} {{n \cdot \left( {n + 1} \right)}}$ and we are done $\displaystyle \square$

Note that you may generalise this, consider $\displaystyle m\in\mathbb{N}$ then: $\displaystyle \sum_{k=0}^{\infty}{\tfrac{\binom{m}{k}}{\binom{n+ k}{k}}\cdot (-1)^k}= \tfrac{n} {{m + n}}$ for all $\displaystyle n\in{\mathbb{Z}^+}$