Question #38962

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!

#"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.#