<s>[INST] How to find the sum of an alternating series? 
Find whether the series diverges and its sum:
  $$\sum_{n = 1}^\infty (-1)^{n+1} \frac{3}{5^n}.$$

I found that the series converges using the Alternating Series test because the absolute value of each $n$ decreases while the value of $n$ increases. Then I took the limit as $x$ approaches infinity of $3/5^x$ which is $0$. 
But I am not sure how to go about finding the sum at this point. This is a practice exam, so the solution is more important then the actual answer.
 [/INST] Notice that$$(-1)^{n+1}\frac{3}{5^n}=-3\frac{(-1)^{n}}{5^{n}}=-3\left(\frac{-1}{5}\right)^{n}$$
Since $\sum_{k=1}^{\infty}ar^{k}=\frac{ar}{1-r}$ (iff $|r|<1$), $$\sum_{n=1}^{\infty}-3\left(\frac{-1}{5}\right)^{n}=\frac{-3\cdot\frac{-1}{5}}{1-\frac{-1}{5}}=\frac{\frac{3}{5}}{\frac{6}{5}}=\frac{1}{2}$$ and the sum converges because $\left|\frac{-1}{5}\right|=\frac{1}{5}<1$
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