Question

Question #38962

Answer 1

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.
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That's it!

Answer 2

`"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.`