Chapter 1: Q7MP (page 1)

Test for convergence: $\sum_{n = 2}^{\infty} \frac{1}{n l n {(n)}^{3}}$

Short Answer

Expert verified

The series $\sum_{n = 2}^{\infty} \frac{1}{n 1 n {(n)}^{3}}$ diverges.

Step by step solution

Concept used to show that the series diverges

Integral test for convergence condition is used for given series which is expressed as follows:

For convergence:

If the function continues, decreasing and positive function then,

If $\int_{k}^{\infty} f (x) d x$ is convergent so is $\sum_{n = k}^{\infty} a_{n}$ .

If $\int_{k}^{\infty} f (x) d x$ is divergent so is $\sum_{n = k}^{\infty} a_{n}$ .

Calculation to show that the series∑n=2∞1n1n(n)3 diverges

Let the function $f (x)$ as $\frac{1}{x \ln (x^{3})}$ . For $x > 0$ the function is positive and continues.

Now, integrate and apply the limit to the function as follows:

$\int_{2}^{\infty} f (x) d x = l i m_{t \to \infty} \int_{2}^{t} \frac{1}{x \ln (x^{3})} d x$

$\int_{2}^{\infty} f (x) d x = l i m_{t \to \infty} \int_{2}^{t} \frac{1}{3 x \ln (x)} d x$

$\ln (x) = p$

Now substitute the value in the above equation as follows:

$\int_{2}^{\infty} f (x) d x = l i m_{t \to \infty} \int_{2}^{t} \frac{1}{3 p} d p = l i m_{t \to \infty} {(\frac{1}{3} \ln p)}_{2}^{t} = l i m_{t \to \infty} {(\frac{1}{3} \ln (\ln (x)))}_{2}^{t} = l i m_{t \to \infty} \frac{1}{3} (\ln (\ln t)) - (\ln (\ln (2))) = l i m_{t \to \infty} \frac{1}{3} (\ln (\ln \infty)) - (\ln (\ln (2))) = \infty \ endgathered$

Thus, the series diverges.

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Test for convergence: $\sum_{n = 2}^{\infty} \frac{1}{n l n {(n)}^{3}}$

Short Answer

Step by step solution

Concept used to show that the series diverges

Calculation to show that the series∑n=2∞1n1n(n)3 diverges

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Test for convergence:∑n=2∞1nln(n)3

Short Answer

Step by step solution

Concept used to show that the series diverges

Calculation to show that the series∑n=2∞1n1n(n)3 diverges

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism

Electric Charge Field and Potential

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Study anywhere. Anytime. Across all devices.

Test for convergence: $\sum_{n = 2}^{\infty} \frac{1}{n l n {(n)}^{3}}$