Question

Determine whether the series $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{3^n}$converges or diverges. Give the sum of the convergent series.

Short answer

The series $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{3^n}$ converges to $\frac{1}{4}$.

Step-by-step solution

Step 1. Given information.

Given a series $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{3^n}$.

Step 2. Find if the series converges or not.

The index starts with 1, rather than 0.

Note that the convergence of a series depends not upon the first few terms but only upon the tail of the series.

The standard form of geometric series is $\sum _{k=0}^{\infty }c{r}^k$.

Here, the series $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{3^n}$has $c=\frac{1}{3}$and $r=-\frac{1}{3}$.

The geometric series converges if and only if $\left|r\right|<1$.

Since $r=-\frac{1}{3}$, it follows that the series localid="1648883874265" $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{3^n}$converges.

Step 3. Find the value to which the series converges.

If the geometric series $\sum _{k=0}^{\infty }c{r}^k$converges, it converges to $\frac{c}{1-r}$.

So, the series $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{3^n}$converges to $\frac{{\displaystyle \frac{1}{3}}}{1-\left(-{\displaystyle \frac{1}{3}}\right)}$, that is $\frac{1}{4}$.