Question

Determine whether the series $\sum _{n=4}^{\infty }\frac{4}{{11}^n}$converges or diverges. Give the sum of the convergent series.

Short answer

The series $\sum _{n=4}^{\infty }\frac{4}{{11}^n}$converges to $\frac{2}{6655}$.

Step-by-step solution

Step 1. Given information.

Given a series $\sum _{n=4}^{\infty }\frac{4}{{11}^n}$.

Step 2. Find if the series converges or not.

The index starts with 2, 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 a geometric series is $\sum _{k=0}^{\infty }c{r}^k$.

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

Here, the series $\sum _{n=4}^{\infty }\frac{4}{{11}^n}$has $c=\frac{4}{{11}^4}$and$r=\frac{1}{11}$.

Since $r=\frac{1}{11}$, it follows that the series localid="1648886017183" $\sum _{n=4}^{\infty }\frac{4}{{11}^n}$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 $\sum _{n=4}^{\infty }\frac{4}{{11}^n}$converges to $\frac{{\displaystyle \frac{4}{{11}^4}}}{1-{\displaystyle \frac{1}{11}}}$, that is $\frac{4}{1331}$or equivalently $\frac{2}{6655}$.