Q. 12 - Calculus

Question

Use Exercise 11 to explain why the root test will be inconclusive for every series $\sum _{k=1}^{\infty }{a}_k\text{ in which }{a}_k$ is a power function .

Step-by-Step Solution

Answer

Hence proved that the root test will be inconclusive because the value of $L=1$ and root test is conclusive at $L=1$.

Step 1. Given Information.

We are given $\sum _{k=1}^{\infty }{a}_k.$

Step 2. Root test.

Now, using Root test,

$$\begin{gathered} \underset{k\to \infty }{\lim }{\left(A{x}^r\right)}^{\frac{1}{k}}=\underset{k\to \infty }{\lim }(A{)}^{\frac{1}{x}}\underset{k\to \infty }{\lim }{\left(x^r\right)}^{\frac{1}{k}} \\ =\underset{k\to \infty }{\lim }(A{)}^{\frac{1}{k}}\underset{k\to \infty }{\lim }\left(x^{\frac{r}{k}}\right) \\ \text{ Since, }{a}^x=e^{\ln \left(a^x\right)}\text{, therefore, } \\ {x}^{\frac{r}{k}}=e^{\ln \left(x^{\frac{r}{2}}\right)} \\ =e^{\frac{r}{k}\ln \left(x\right)} \\ Now, \\ \underset{k\to \infty }{\lim }\left(x^{\frac{r}{k}}\right)=\underset{k\to \infty }{\lim }{e}^{\frac{r}{k}\ln \left(x\right)} \\ =e^{\underset{i\to \infty }{\lim }\left(\frac{r}{k}\ln \left(x\right)\right)} \\ =1 \end{gathered}$$

Therefore,

$$\begin{gathered} \underset{k\to \infty }{\lim }{\left(A{x}^r\right)}^{\frac{1}{k}}=1 \\ \underset{k\to \infty }{\lim }{a}_k{}^{\frac{1}{k}}=1 \\ L=1 \\ Therefore, it is a inconclusive. \end{gathered}$$