Chapter 7: Q. 11 (page 631)

Use the integral test to show that the series $\sum_{k = 2}^{\infty} \frac{1}{k \ln k}$ diverges.

Short Answer

Expert verified

The series $\sum_{k = 2}^{\infty} \frac{1}{k \ln k}$ is divergent.

Step by step solution

Step 1. Given information

A series is given as $\sum_{k = 2}^{\infty} \frac{1}{k \ln k}$

Step 2. Integral

Here all conditions of integral such as continuous, positive terms, etc. are true for the given series. So we can do integration.

The integral is $\int_{x = 2}^{\infty} f (x) d x = \int_{x = 2}^{\infty} \frac{1}{x (\ln x)} d x$

Now simplify it as,

$\int_{x = 2}^{\infty} \frac{1}{x (\ln x)^{2}} d x = \lim_{i \to \infty} \int_{x = 2}^{1} \frac{1}{x (\ln x)} d x = \lim_{k \to \infty} \int_{u = \ln 2}^{\ln k} \frac{1}{u} d u (Put \ln x = u, \Rightarrow \frac{1}{x} d x = d u) = \lim_{k \to \infty} [\ln u]_{\ln 2}^{\ln k} = \lim_{k \to \infty} [\ln (\ln k) - \ln (\ln 2) = \infty$

Hence the series is divergent.

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Use the integral test to show that the series $\sum_{k = 2}^{\infty} \frac{1}{k \ln k}$ diverges.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Integral

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Use the integral test to show that the series∑k=2∞1klnkdiverges.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Integral

One App. One Place for Learning.

Most popular questions from this chapter

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Theoretical and Mathematical Physics

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Use the integral test to show that the series $\sum_{k = 2}^{\infty} \frac{1}{k \ln k}$ diverges.