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.
Best Answer
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 $\left|\frac{-1}{5}\right|=\frac{1}{5}<1$