Answer
convergent;
$S_{\infty}=\dfrac{2}{3}$
Work Step by Step
RECALL:
(1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula:
$S_{\infty}=\dfrac{a}{1-r}$
(2) An infinite geometric series is divergent if $|r|>1$.
(3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it:
$r = \dfrac{a_n}{a_{n-1}}$
Solve for $r$ to obtain:
$\require{cancel}
r = \dfrac{\frac{4}{25}}{\frac{2}{5}}
\\r=\dfrac{4}{25} \cdot \dfrac{5}{2}
\\r=\dfrac{\cancel{4}2}{\cancel{25}5} \cdot \dfrac{\cancel{5}}{\cancel{2}}
\\r=\dfrac{2}{5}$
Since $|\frac{2}{5}|<1$, then the series converges.
Solve for the sum using the formula above, with $a=\dfrac{2}{5}$ and $r=\frac{2}{5}$, to obtain:
$\require{cancel}
S_{\infty}=\dfrac{\frac{2}{5}}{1-\frac{2}{5}}
\\S_{\infty}=\dfrac{\frac{2}{5}}{\frac{5}{5}-\frac{2}{5}}
\\S_{\infty}=\dfrac{\frac{2}{5}}{\frac{3}{5}}
\\S_{\infty} = \dfrac{2}{5} \cdot \frac{5}{3}
\\S_{\infty} = \dfrac{2}{\cancel{5}} \cdot \dfrac{\cancel{5}}{3}
\\S_{\infty}=\dfrac{2}{3}$
