"Let's look a little further into the series. We have:"
sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \ = \ sum_{n=1}^{infty} \ 3^{n} cdot 3^2 cdot 5^{-n}
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 3^2 cdot sum_{n=1}^{infty} \ 3^{n} cdot 5^{-n}
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 3^2 cdot sum_{n=1}^{infty} \ 3^{n}/ 5^{n}
\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 3^2 cdot sum_{n=1}^{infty} \ ( 3/5 )^{n}.
"Thus:"
\qquad \qquad \qquad \qquad \qquad \quad sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \ = \ 3^2 cdot sum_{n=1}^{infty} \ ( 3/5 )^{n}. \qquad \qquad \qquad \qquad \ \ \ (1)
"Now we see that the series we have in the RHS of (1) is an"
"infinite geometric series, with common ratio" \ \ r = 3/5. \ \ "As" \ \ |r| < 1, "the series converges. And, also by eqn. (1), we see the"
"series you gave originally is a constant multiple of this infinite"
"geometric series [in fact, a multiple of" \ \ 3^2 \ "of it"]. "So the series"
"you gave originally, clearly converges, too."
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \qquad \qquad "converges."
"While not asked, it is hard to resist the fact that, for infinite"
"geometric series, with common ratio having absolute value less"
"than 1, there is a nice formula for the sum of the infinite series."
"Recall that:"
\qquad \qquad \qquad \qquad \qquad \qquad \quad |r| < 1 \quad => \quad sum_{k=0}^{infty} a r^k \ = \ a/{ 1 - r } \qquad \qquad \qquad \qquad \quad \quad \quad \ (2)
"Continuing from eqn. (1), and using eqn. (2) with eqn. (1), we"
"have:"
\qquad \qquad \ \ sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \ = \ 3^2 cdot sum_{n=1}^{infty} \ ( 3/5 )^{n} \qquad \qquad \qquad \qquad \ \ \
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \ = \ 3^2 cdot [ sum_{n=0}^{infty} \ ( 3/5 )^{n} - (3/5)^0]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \ = \ 3^2 cdot [ sum_{n=0}^{infty} \ ( 3/5 )^{n} - 1 ]
\qquad "using eqn. (2):" \qquad \quad \ \ = \ 3^2 cdot [ ( 1/{1 -
3/5} ) - 1 ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \ = \ 3^2 cdot [ 5/{5 - 3} - 1 ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \ = \ 3^2 cdot [ 5/2 - 1 ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \ = \ 3^2 cdot [ 3/2 ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \ = \ 27/2.
"Thus:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \ = \ 27/2 \quad.