Convergence of $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + ...$ How does one use the comparison test to prove that  $ 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$ converges?
Is the following argument valid?
$\quad 1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$
$< 1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^3} + \;... $
$= 2 + 1 + \frac{1}{2} + \frac{1}{2^2} + \;... $
$= \frac{2}{1-1/2} = 4 $
So the series is bounded above and since every term is strictly positive, the sequence of partial sums is monotone increasing and hence the series converges.

$$S=1+1+\frac{1}{3} + \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{2^3} + \; ...$$
$$=\sum_{0\le r<\infty} \left(\frac12\right)^r+\sum_{0\le r<\infty} \left(\frac13\right)^r$$
Using summation formula of Infinite Geometric Series
$$S=\frac1{1-\frac12}+\frac1{1-\frac13}$$
