Chapter 1: Q14P (page 1)

Question: Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.
14. $\sum_{1}^{\infty} \frac{1}{\sqrt{n^{2} + 9}}$

Short Answer

Expert verified

The series $\sum_{1}^{\infty} \frac{1}{\sqrt{n^{2} + 9}}$ is divergent.

Step by step solution

Definition of convergent and divergent

If the partial sum $S_{n}$ of an infinite series tend to a limit S, then the series is called convergent.

If the partial sum $S_{n}$ of an infinite series do not approach to a limit, the series is called divergent.

The limiting value S is called the sum of the series.

Apply the integral test

The given series is $\sum_{1}^{\infty} \frac{1}{\sqrt{n^{2} + 9}}$ .

Use the integral in the given series as:

$\int_{}^{\infty} \frac{1}{\sqrt{n^{2} + 9}} d n$

Now, solve the integral is as follows:

Use the integral formula $\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \log |x + \sqrt{x^{2} + a^{2}}| + c$ , where c is a constant.

$\begin{array}{rcl} \int_{}^{\infty} \frac{1}{\sqrt{n^{2} + 9}} d n & = & {[\log |n + \sqrt{n^{2} + 9}|]}^{\infty} \\ = & [\log |\infty + \sqrt{\infty + 9}|] \\ = & \infty \end{array}$

Here, the series approaches to infinite, therefore the given series diverges.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Question: Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.
14. $\sum_{1}^{\infty} \frac{1}{\sqrt{n^{2} + 9}}$

Short Answer

Step by step solution

Definition of convergent and divergent

Apply the integral test

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Turning Points in Physics

Mechanics and Materials

Measurements

Translational Dynamics

Force

Study anywhere. Anytime. Across all devices.

Question: Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.14.∑1∞1n2+9

Short Answer

Step by step solution

Definition of convergent and divergent

Apply the integral test

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Turning Points in Physics

Mechanics and Materials

Measurements

Translational Dynamics

Force

Study anywhere. Anytime. Across all devices.

Question: Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.
14. $\sum_{1}^{\infty} \frac{1}{\sqrt{n^{2} + 9}}$