Convergence of alternating series with non decreasing terms I need to test the following alternating series for convergence: 
$$\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{2^{4}}-\frac{1}{3^{4}}+\dots$$
Here is my thinking: The series is divergent because by the "test for alternating series" the terms do not decrease. For instance $\frac{1}{3^{4}}<\frac{1}{2^{5}}$. 
However, Boas (Mathematical Methods in the physical sciences, chap. 1 section 9 question 20) suggests that this series is convergent. Why is that? 
 A: You have the geometric sum $$\sum_{n=2}^\infty\left(\frac{1}{2^n}-\frac{1}{3^n}\right) = \sum_{n=2}^\infty\frac1{2^n}-\sum_{n=2}^\infty\frac{1}{3^n} = \frac{1/{2^2}}{1-1/2} - \frac{1/{3^2}}{1-1/3} = \frac13,$$
since $\sum_{n=a}^\infty r^n = \frac{r^a}{1-r}$ for $|r|<1$.
A: Yes, your sum converges and indeed its value is 
$$
\sum_{n=2}^{\infty} \left( \frac{1}{2^n} - \frac{1}{3^n}\right) = \frac13.
$$
This is easy to see as $\displaystyle \sum_{n=2}^{\infty} \frac{1}{2^n} = \frac12$ and $\displaystyle  \sum_{n=2}^{\infty} \frac{1}{3^n} = \frac16$.
A: Consider the two series $\sum_{k=0}^{\infty} 2^{-k} = 2$ and $\sum_{k=0}^{\infty} 3^{-k} = \dfrac{3}{2}$. Then, your series is just 
$$\sum_{k=2}^{\infty} (2^{-k} - 3^{-k}) = \sum_{k=0}^{\infty} 2^{-k} - \sum_{k=0}^{\infty} 3^{-k} -\dfrac{1}{6}= \dfrac{1}{3}.$$
You can of course ask, whether it is absolute convergent but this follows from $2^{-k}\geq 3^{-k}$ for all $k\geq 0$ such that all addends are positive.
EDIT: sorry, I had a little error in calculations.
A: Hint:
$$\sum_{n=2}^{\infty} \left( \frac{1}{2^n} - \frac{1}{3^n}\right)  =\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{2^n}-\frac{1}{9}\sum_{n=0}^{\infty}\frac{1}{3^n} $$
