# Find z such that 1 +\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...= 3.

## Question:

Find {eq}z {/eq} such that {eq}1 +\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...= 3. {/eq}

## Binomial Series and its Specific case:

If {eq}k {/eq} is any number and {eq}\left| x \right| < 1 {/eq} then Binomial series expansion is given as:

{eq}{(1 + x)^k} = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} k \\ n \\ \end{array}} \right){x^n}} = 1 + kx + \frac{{k(k - 1)}}{{2!}}{x^2} + \frac{{k(k - 1)(k - 2)}}{{3!}}{x^3} + ..... {/eq}

Also we know that:

{eq}\eqalign{ & {\left( {1 - x} \right)^{ - 1}} = 1 + x + {x^2} + {x^3} + .... \cr & {\left( {1 + x} \right)^{ - 1}} = 1 - x + {x^2} - {x^3} + .... \cr} {/eq}

Using Binomial expansion the given equation can be represented as:

{eq}\eqalign{ 1 + \frac{1}{z} + \frac{1}{{{z^2}}} + \frac{1}{{{z^3}}} + ... = 3 & \Rightarrow 1 + \frac{1}{z} + {\left( {\frac{1}{z}} \right)^2} + {\left( {\frac{1}{z}} \right)^3} + ... = 3 \cr & \Rightarrow {\left( {1 - \frac{1}{z}} \right)^{ - 1}} = 3 \cr & \Rightarrow {\left( {\frac{{z - 1}}{z}} \right)^{ - 1}} = 3 \cr & \Rightarrow \left( {\frac{z}{{z - 1}}} \right) = 3 \cr & \Rightarrow z = 3z - 3 \cr & \Rightarrow - 2z = - 3 \cr & \Rightarrow 2z = 3 \cr & \Rightarrow z = \frac{3}{2} \cr} {/eq}