Question

Determine whether the series $\sum _{k=0}^{\infty }\mathrm{\pi }{\left(\frac{\mathrm{e}}{3}\right)}^{\mathrm{k}}$converges or diverges. Give the sum of the convergent series.

Short answer

The series $\sum _{k=0}^{\infty }\mathrm{\pi }{\left(\frac{\mathrm{e}}{3}\right)}^{\mathrm{k}}$ converges to $\frac{3\mathrm{\pi }}{3-e}$.

Step-by-step solution

Step 1. Given information.

Given a series $\sum _{k=0}^{\infty }\mathrm{\pi }{\left(\frac{\mathrm{e}}{3}\right)}^{\mathrm{k}}$.

Step 2. Find if the series converges or not.

The series $\sum _{k=0}^{\infty }\mathrm{\pi }{\left(\frac{\mathrm{e}}{3}\right)}^{\mathrm{k}}$is in the standard form $\sum _{k=0}^{\infty }c{r}^k$for a geometric series with $c=\mathrm{\pi }$and $r=\frac{e}{3}$.

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

Note that $2<e<3$. It follows that $\frac{e}{3}<1$.

Therefore, the series localid="1648958893904" $\sum _{k=0}^{\infty }\mathrm{\pi }{\left(\frac{\mathrm{e}}{3}\right)}^{\mathrm{k}}$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 $\sum _{k=0}^{\infty }\mathrm{\pi }{\left(\frac{\mathrm{e}}{3}\right)}^{\mathrm{k}}$converges to $\frac{\mathrm{\pi }}{1-{\displaystyle \frac{e}{3}}}$, that is $\frac{3\mathrm{\pi }}{3-e}$.