After solving a probability question I ended up with the following:
$P(A)=\frac{1}{6}\sum\limits_{n=1}^\infty(\frac{5}{6})^{3n-2}$ limit between 1 and infinity but I'm unsure how to carry this on , I am familiar with summing $x^n$ to infinity for modulus X less than one . Sorry if this is a easy question, any help would be great thanks.
You may write $$ \begin{align} \frac16\sum_{x=1}^\infty \left(\frac{5}{6}\right)^{3x-2}&= \frac16 \cdot \left(\frac{5}{6}\right)^{-2}\sum_{x=1}^\infty \left(\frac{5}{6}\right)^{3x} \\\\&= \frac16 \cdot \frac{36}{25}\sum_{x=1}^\infty \left(\frac{5^3}{6^3}\right)^{x} \\\\&=\frac{6}{25}\sum_{x=1}^\infty \left(\frac{125}{216}\right)^{x} \\\\&=\frac{6}{25}\cdot\frac{125}{216} \sum_{x=1}^\infty \left(\frac{125}{216}\right)^{x-1} \\\\&=\frac{5}{36} \sum_{x=0}^\infty \left(\frac{125}{216}\right)^{x} \end{align} $$ by a change of index.
Hope you can finish it using $$ \sum_{x=0}^\infty q^{x}=\frac 1{1-q},\quad|q|<1. $$
I will use the fact that $\sum\limits_{n=0}^\infty p^n=\frac{1}{1-p}$ if $|p|<1$.
Now,
$\begin{align*} \frac{1}{6}\sum\limits_{n=1}^\infty (\frac{5}{6})^{3n-2} &= \frac{1}{6}\left(\sum\limits_{n=1}^\infty (\frac{5}{6})^{3n}\right)(\frac{5}{6})^{-2}\\ &= \frac{6}{25} \left(\sum\limits_{n=1}^\infty [(\frac{5}{6})^3]^{n}\right)\\ \end{align*}$
Now put $p=(\frac{5}{6})^3$.
We have $\frac{6}{25}\sum\limits_{n=1}^\infty p^n=\frac{6}{25}\left(\sum\limits_{n=0}^\infty p^n-p^0\right)=\frac{6}{25}\left(\frac{1}{1-p}-1\right)=\frac{6}{25}\frac{p}{1-p}=\frac{6}{25}\frac{(\frac{5}{6})^3}{1-(\frac{5}{6})^3}$
It is a decreasing geometric progression with $q=\left(\frac56\right)^3<1$: $$\frac16\sum_{x=1}^\infty \left(\frac{5}{6}\right)^{3x-2}=\frac16\left(\frac56+\left(\frac56\right)^4+\left(\frac56\right)^7+\cdots\right)=\frac16\cdot \frac{\frac56}{1-\left(\frac56\right)^3}=\frac{30}{91}.$$
