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?
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4 Answers
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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 $|r|<1$.
answered Jul 5, 2019 at 9:42
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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 $\displaystyle \sum_{n=2}^{\infty} \frac{1}{3^n} = \frac16$.
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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.
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Hint: $$\sum_{n=2}^{\infty} \left( \frac{1}{2^n} - \frac{1}{3^n}\right) =\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{2^n}-\frac{1}{9}\sum_{n=0}^{\infty}\frac{1}{3^n} $$
