The [tex]n[/tex]-th term in the series is 6 multiplied by the [tex](n-1)[/tex]-th power of 5/6:
[tex]a_1=6=6\left(\dfrac56\right)^{1-1}[/tex]
[tex]a_2=5=6\left(\dfrac56\right)^{2-1}[/tex]
[tex]a_3=\dfrac{25}6=6\left(\dfrac56\right)^{3-1}[/tex]
and so on.
[tex]\displaystyle\sum_{n=1}^\infty6\left(\frac56\right)^{n-1}[/tex]
Consider the [tex]N[/tex]-th partial sum,
[tex]S_N=\displaystyle\sum_{n=1}^N6\left(\frac56\right)^{n-1}[/tex]
[tex]S_N=6\left(1+\dfrac56+\cdots+\dfrac{5^{N-2}}{6^{N-2}}+\dfrac{5^{N-1}}{6^{N-1}}\right)[/tex]
Multiplying both sides by 5/6 gives
[tex]\dfrac56S_N=6\left(\dfrac56+\dfrac{5^2}{6^2}+\cdots+\dfrac{5^{N-1}}{6^{N-1}}+\dfrac{5^N}{6^N}\right)[/tex]
and substracting this from [tex]S_N[/tex] gives
[tex]\dfrac16S_N=6\left(1-\dfrac{5^N}{6^N}\right)[/tex]
[tex]S_N=36\left(1-\left(\dfrac56\right)^N}\right)[/tex]
As [tex]N\to\infty[/tex], it's clear that the sum converges to 36.
