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.
 A: $$\underset{n\geq1}{\sum}\left(-1\right)^{n+1}\frac{1}{3^{n}}=-\underset{n\geq1}{\sum}\left(\frac{-1}{3}\right)^{n}=-\left(\underset{n\geq0}{\sum}\left(\frac{-1}{3}\right)^{n}-1\right)=\frac{1}{4}.$$Note that the geometric series starts from $0$.
A: First, determine the general term $$a_n=-\left(\frac{-1}{3}\right)^n$$
Next, it is often a good idea to calculate $\frac{a_n}{a_{n-1}}$.  In this case you get $$\frac{a_n}{a_{n-1}}=-\frac{1}{3}$$
Now, if this is a constant, such as in this case, you have a geometric series.  If not, you can take its limit as $n\to \infty$ and use the ratio test to determine convergence (unless the limit is $1$).
A: $\sum_{n=0}^\infty(-1)^{n+1} \cdot \frac {1}{3^n} $ is sum 
$-\sum_{n=0}^{\infty}a^n $ where $a=\left(\frac{-1}{3}\right)^n$
In theory of sum function that correspond to this sum is f(a)=-1/(1-a)+1.
Abel thm implies that sum  is f(-1/3)=-3/4.
You see here for more detail: http://mathworld.wolfram.com/MaclaurinSeries.html
Yes, seris start from 1 than you have: sum = -3/4+1=1/4
