Q.17 - Calculus

Question

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

Step-by-Step Solution

Answer

the root test is used to prove the series converges

Step 1: Given information

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.

Step 2: Calculation

Now according to convergence and divergence of geometric series,

 

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

Step 3: Further Calculation

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

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 $\frac{1}{1-r}$.

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

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

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