Chapter 7: Q. 12 (page 639)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

We are given $\sum_{k = 1}^{\infty} a_{k} .$

Step 2. Root test.

Now, using Root test,

$\lim_{k \to \infty} {(A x^{r})}^{\frac{1}{k}} = \lim_{k \to \infty} (A)^{\frac{1}{x}} \lim_{k \to \infty} {(x^{r})}^{\frac{1}{k}} = \lim_{k \to \infty} (A)^{\frac{1}{k}} \lim_{k \to \infty} (x^{\frac{r}{k}}) Since, a^{x} = e^{\ln (a^{x})}, therefore, x^{\frac{r}{k}} = e^{\ln (x^{\frac{r}{2}})} = e^{\frac{r}{k} \ln (x)} N o w, \lim_{k \to \infty} (x^{\frac{r}{k}}) = \lim_{k \to \infty} e^{\frac{r}{k} \ln (x)} = e^{\lim_{i \to \infty} (\frac{r}{k} \ln (x))} = 1$

Therefore,

$\lim_{k \to \infty} {(A x^{r})}^{\frac{1}{k}} = 1 \lim_{k \to \infty} a_{k} {}^{\frac{1}{k}}= 1 L = 1 T h e r e f o r e, i t i s a i n c o n c l u s i v e .$

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

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Root test.

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Use Exercise 11 to explain why the root test will be inconclusive for every series ∑k=1∞akin whichakis a power function .

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Root test.

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

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