explain why the series$\sum _{k=0}^{\infty }\frac{1}{3^k}$converges. which convergence tests could be used to prove this?

Consider the series $\sum _{k=0}^{\infty }\frac{1}{3^k}$

To explain why the series and which test is used to prove the series converges.

The series $\sum _{k=0}^{\infty }\frac{1}{3^k}$is of the form $\sum _{k=0}^{\infty }{r}^k$, where r is a non-zero real number.

Now according to convergence and divergence of geometric series,

1. For $\left|r\right|<1$,the geometric series localid="1651557564654" $\sum _{k=0}^{\infty }{r}^k$converges to the sum $\frac{1}{1-r}$.

2. For $\left|r\right|\ge 1$, the geometric series $\sum _{k=0}^{\infty }{r}^k$diverges.

Now, the series $\sum _{k=0}^{\infty }\frac{1}{3^k}$is of the form localid="1661335365334" $\sum _{k=0}^{\infty }{r}^k,\mathrm{where}r=\frac{1}{3}.$

Now,$\left|r\right|$<1 , therefore, the series $\sum _{k=0}^{\infty }\frac{1}{3^k}$converges and also the series converges to the sum localid="1661335395783" $\frac{1}{1-r}$.

Calculate the value of,$\frac{1}{1-r}$.

localid="1661335430032" $\begin{array}{rcl}\frac{1}{1-r}& =& \frac{1}{1-{\displaystyle \frac{1}{3}}}\\ & =& \frac{1}{{\displaystyle \frac{2}{3}}}\\ & =& \frac{3}{2}\end{array}$

Hence the series converges to $\frac{3}{2}$and the test which could be used to test the convergence test is the root test.