Limit and Ratio test

Question

Here the problem 2.3.29(5) from Sohrab, Basic Real Analysis

Problem

Investigate the convergence or divergence of the following series $\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots$

Hint: note that the Ratio Test is inconclusive.

Solution

$x_k=\frac{1}{2^k}+\frac{1}{3^k}$. Ratio test:

$$ \lim(x_{n+1}/x_n)=\lim\frac{\frac{1}{2^{k+1}}+\frac{1}{3^{k+1}}}{\frac{1}{2^k}+\frac{1}{3^k}}=\lim \frac{1}{6}\frac{3^{k+1}+2^{k+1}}{3^k+2^k}\\ =\lim\frac{1}{6}\frac{1+(\frac{2}{3})^{k+1}}{\frac{1}{3}(1+(\frac{2}{3})^k)}=\frac{1}{2}<1. $$

Thus $\sum(\frac{1}{2^k}+\frac{1}{3^k})$ is convergent.

Question

Is it possible that the hint is wrong? Otherwise, what is my error?

Answer 1

There are multiple ways to write the given expression precisely as an infinite series. I think the problem-writer's thinking was to write it as $\sum_{k=1}^\infty x_k$ where $$ x_k = \begin{cases} \frac1{2^{(k+1)/2}}\,, &\text{if $k$ is odd}, \\ \frac1{3^{k/2}}, &\text{if $k$ is even}. \end{cases} $$ In this form, the Ratio Test is indeed inconclusive. However, if you rewrite it in the form you gave, than the Ratio Test is fine. Since all the terms are positive, you could also just separate the powers of $\frac12$ from the powers of $\frac13$ (into two infinite series) and evaluate each one individually.