- #1

- 140

- 0

## Homework Statement

Consider the series

[tex]\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \ldots[/tex].

He then proceeds to write

[tex]\lim \textnormal{inf}_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left(\frac{2}{3}\right)^n = 0[/tex]

and

[tex]\lim\textnormal{sup}_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{1}{2} \left(\frac{3}{2}\right)^n = +\infty[/tex].

This is in reference to the ratio test, by the way.

## Homework Equations

## The Attempt at a Solution

How did he arrive at these calculations? If for the lower limit we select [itex]a_{n+1} = 1/3^n[/itex] and [itex]a_n = 1/2^n[/itex], then we can recover his answer. But

[tex]\left\{\frac{a_{n+1}}{a_n}\right\} = \left\{\left(\frac{2}{3}\right)^n\right\}[/tex]

is a convergent sequence and the upper and lower limits must be equal, correct? How is the upper limit [itex]\infty[/itex]?

On the other hand, if we select [itex]a_{n+1} = 1/2^{n+1}[/itex] and [itex]a_n = 1/3^n[/itex], then we can recover his answer for the upper limit.

Isn't there some ambiguity with selecting [itex]a_{n+1}[/itex] and [itex]a_n[/itex] from this series?