The terms in the sum are given by the sequence
[tex]\left\{\dfrac{5^{n+1}}{3^n}\right\}_{n\ge1}[/tex]
We have
[tex]\dfrac{5^{n+1}}{3^n}>\dfrac{5^n}{3^n}=\left(\dfrac53\right)^n>1[/tex]
for all [tex]n[/tex], and the series [tex]1+1+1+\cdots[/tex] clearly diverges, so [tex]\dfrac{25}3+\dfrac{125}9+\cdots[/tex] must also diverge.
