Question

Determine whether the geometric series is convergent or divergent. 6 + 5 + 25/6 + 125/36 + ...

Answer 1

The $n$-th term in the series is 6 multiplied by the $(n-1)$-th power of 5/6:

$a_1=6=6\left(\dfrac56\right)^{1-1}$

$a_2=5=6\left(\dfrac56\right)^{2-1}$

$a_3=\dfrac{25}6=6\left(\dfrac56\right)^{3-1}$

and so on.

$\displaystyle\sum_{n=1}^\infty6\left(\frac56\right)^{n-1}$

Consider the $N$-th partial sum,

$S_N=\displaystyle\sum_{n=1}^N6\left(\frac56\right)^{n-1}$

$S_N=6\left(1+\dfrac56+\cdots+\dfrac{5^{N-2}}{6^{N-2}}+\dfrac{5^{N-1}}{6^{N-1}}\right)$

Multiplying both sides by 5/6 gives

$\dfrac56S_N=6\left(\dfrac56+\dfrac{5^2}{6^2}+\cdots+\dfrac{5^{N-1}}{6^{N-1}}+\dfrac{5^N}{6^N}\right)$

and substracting this from $S_N$ gives

$\dfrac16S_N=6\left(1-\dfrac{5^N}{6^N}\right)$

$S_N=36\left(1-\left(\dfrac56\right)^N}\right)$

As $N\to\infty$, it's clear that the sum converges to 36.