Value and nature of $\sum_{n=1}^\infty(-1)^{n+1} \cdot \frac {1}{3^n}$ $$\sum_{n=1}^\infty(-1)^{n+1} \cdot \frac {1}{3^n}$$ Find the sum and determine it's nature.
I have tried to find the limit of partial sums, but I could not do anything. There is another bunch of exercises of this type in front of me; can you please show me in detail how this exercise can be solved? Thank you.

$$\underset{n\geq1}{\sum}\left(-1\right)^{n+1}\frac{1}{3^{n}}=-\underset{n\geq1}{\sum}\left(\frac{-1}{3}\right)^{n}=-\left(\underset{n\geq0}{\sum}\left(\frac{-1}{3}\right)^{n}-1\right)=\frac{1}{4}.$$Note that the geometric series starts from $0$.
