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Chapter 7: Q. 25 (page 657)

Check the convergence $\sum_{k = 2}^{\infty} \frac{1}{\sqrt[2]{e}}$

Short Answer

Expert verified

Diverges.

Step by step solution

01

Step 1. Given

The given series is $\sum_{k = 2}^{\infty} \frac{1}{\sqrt[2]{e}}$ .

02

Step 2. Divergence test

According to the divergence test if the sequence {a_k} does not converge to zero then the series $\sum_{k = 1}^{\infty} a_{k}$ diverges .

03

Step 3. Checking the convergence

$Consider the sequence a_{k} a_{k} = \frac{1}{\sqrt[k]{e}} = \frac{1}{e^{\frac{1}{k}}} The value of \lim_{k \to \infty} a_{k} i s \lim_{k \to \infty} a_{k} = \lim_{k \to \infty} \frac{1}{e^{\frac{1}{k}}} = 1 Since, \lim_{k \to \infty} a_{k} ≅ 0 Here a_{k} does not converge to 0 . Hence, the series diverges .$

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Most popular questions from this chapter

Let 0 < p < 1. Evaluate the limit $\lim_{k \to \infty} \frac{1 / k \ln k}{1 / k^{p}}$

Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series $\sum_{k = 2}^{\infty} \frac{1}{k \log k}$

For a convergent series satisfying the conditions of the integral test, why is every remainder $R_{n}$ positive? How can $R_{n}$ be used along with the term $S_{n}$ from the sequence of partial sums to understand the quality of the approximation $S_{n}$ ?

Express each of the repeating decimals in Exercises 71–78 as a geometric series and as the quotient of two integers reduced to lowest terms.

$0.199999 . . .$

Determine whether the series $\sum_{k = 0}^{\infty} e {(\frac{π}{3})}^{k}$ converges or diverges. Give the sum of the convergent series.

Prove Theorem 7.24 (a). That is, show that if c is a real number and $\sum_{k = 1}^{\infty} a_{k}$ is a convergent series, then $\sum_{k = 1}^{\infty} c a_{k} = c \sum_{k = 1}^{\infty} a_{k}$ .

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