Question #38962

2 Answers
Feb 22, 2018

The given series converges!

Explanation:

If there exists an N so that for all n\ge N, \ \ \ \ \ a_n\ne 0\ \ \ \ \ and\ \ \ \ \ lim_{n\to \infty }|\frac{a_{n+1}}{a_n}|=L

  • If L<1, then \sum a_n \ \ \ \ \text{converges}
  • If L>1, then \sum a_n \ \ \ \ \text{diverges}
  • If L=1, \ \ \text{then the test is inconclusive}

|\frac{a_{n+1}}{a_n}|=|\frac{3^{(n+1)+2}\cdot 5^{-(n+1)}}{3^{n+2}\cdot 5^{-n}}|

Simplify to get:

=\frac{3}{5}

And:

\lim _{n\to \infty }(\frac{3}{5})=\frac{3}{5}

By the geometric ratio test, L<1, so the given series converges.


That's it!

Feb 22, 2018

"a)" \qquad \qquad sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \qquad \qquad "converges."

["b) Not asked for, but connected:" \quad \quad \ sum_{n=1}^{infty} 3^{n+2} cdot 5^{-n} \ = \ 27/2\quad.]

Explanation:

"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.