Question

Determine whether the following series converges or diverges using the integral test. Be sure to verify that the integral test can be applied.

Answer 1

By using the integral test, series converges

What is an integral test for convergence and divergence?

⇒ It is used to prove the divergence or convergence of series. This test is called the integral test, which compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

How to know if a series is converging or diverging?

If the limit exists and is a finite number (a number less than infinity), we say the integral converges. If the limit is ±∞ or does not exist, we say the integral diverges.

let aₓ= f (x) is any function

⇒ In order for the integral test to work The function must be positive it has to be continuous and it has to be decreasing when x≥1

If the integral converges it means the series is also converging

If the integral diverges it means the series is also diverging

The sequence $k^{3} /{e^{k^{4} }$ is clearly positive and decreases for k∈N then by the integral test,

$\int\limits^\infty_1 {x^{3} /e^{x^4} dx$ $\leq\displaystyle \sum^{\infty}_{k = 1} {x^{3} /e^{x^{4}$

and

$\int\limits^\infty_1 {x^{3} /e^{x^4} dx$$=1/4\int\limits^\infty_1 {x^{3} /e^{-u} du$

⇒ $1/4 < \infty$

So it comes with a finite value hence the series converges

Learn more about the integral tests here :

brainly.com/question/15394015

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Answer 2

By the integral test, the series is convergent.

The sequence $k^3/e^{k^4}$ is clearly positive and decreasing for $k\in\Bbb N$; then by the integral test,

$\displaystyle \int_1^\infty \frac{x^3}{e^{x^4}} \, dx \le \sum_{k=1}^\infty \frac{k^3}{e^{k^4}}$

and

$\displaystyle \int_1^\infty \frac{x^3}{e^{x^4}} \, dx = \frac14 \int_1^\infty e^{-u}\, du = \frac1{4e} < \infty$