<s>[INST] 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.
 [/INST] Using the root test, we can prove easily,
note that,
$$\limsup_{n\to \infty} \sqrt[n]a_n=\lim_{n\to \infty}\sqrt[2n]{\frac 1 {2^n}}=\frac 1 {\sqrt 2}<1 $$
Hence the convergence.
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