Find the sum of the series: sum from n=1 to infinity
Find the sum of the series: $$ \sum_{n=1}^{\infty}\left(\frac1{3^n}+\frac{(-1)^n}{2^n}\right). $$
Explanation of Method
Both pieces of the summand are, separately, convergent geometric series (ratios $\frac13$ and $-\frac12$, both of magnitude less than $1$), so the sum can be split term-by-term and each geometric series evaluated by the standard closed form $\sum r^n=\dfrac r{1-r}$ (for $n\ge1$). No convergence test is needed beyond recognizing the geometric structure.
- Split the series into its two geometric components.
- Sum each geometric series using $\sum_{n=1}^\infty r^n=\dfrac r{1-r}$.
- Combine the two results.
Step 1 · Split into two geometric series
Both $\sum\frac1{3^n}$ and $\sum\frac{(-1)^n}{2^n}$ converge absolutely (geometric with $|r|<1$), so linearity of convergent series allows the split: $$ \sum_{n=1}^{\infty}\left(\frac1{3^n}+\frac{(-1)^n}{2^n}\right) = \sum_{n=1}^{\infty}\frac1{3^n} + \sum_{n=1}^{\infty}\frac{(-1)^n}{2^n}. $$
Step 2 · Sum each geometric series
$$ \sum_{n=1}^{\infty}\frac1{3^n} = \frac{1/3}{1-1/3} = \frac{1/3}{2/3} = \frac12. $$
$$ \sum_{n=1}^{\infty}\frac{(-1)^n}{2^n} = \sum_{n=1}^{\infty}\left(-\frac12\right)^n = \frac{-1/2}{1-(-1/2)} = \frac{-1/2}{3/2} = -\frac13. $$
Step 3 · Combine
$$ \sum_{n=1}^{\infty}\left(\frac1{3^n}+\frac{(-1)^n}{2^n}\right) = \frac12 - \frac13 = \frac16. $$
Summary Table
| Component | Value |
|---|---|
| $\sum1/3^n$ | $1/2$ |
| $\sum(-1)^n/2^n$ | $-1/3$ |
| Total | $1/6$ |
Additional Notes
- This is the simplest possible problem in the entire 121–140 range: both component series are elementary geometric series with no oscillating trigonometric factor, no parametrized threshold, and no need for Leibniz's, Dirichlet's, or Abel's tests — a deliberate closing "reset" after the intricate perturbation problems (134, 135) and theorem/counterexample problems (136, 137) that precede it.
- Alongside Problem 139, this is one of only two problems in this batch with a fully closed-form rational answer, both relying on nothing more than the elementary geometric series formula $\sum_{n=1}^\infty r^n=r/(1-r)$ for $|r|<1$.
- The second component, $\sum(-1)^n/2^n=\sum(-1/2)^n$, is itself a special case of Problem 128's parametrized family $\sum(-1)^{n-1}(2\sin^2x)^n/n$ with the $1/n$ factor removed — stripping that factor turns a conditionally-convergent-at-the-boundary family into a purely geometric one, absolutely convergent throughout its entire disk $|r|<1$.
Result
The series converges to: