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How do you use the Integral test on the infinite series #sum_(n=1)^oo1/sqrt(n+4)# ?

Answer 1

Christopher Agresta

Since the integral

#int_1^infty 1/sqrt{x+4} dx#

diverges, the series

#sum_{n=1}^infty1/sqrt{n+4}#

also diverges by Integral Test.

Let us evaluate the integral.

#int_1^infty 1/sqrt{x+4} dx#

by the definition of improper integral,

#=lim_{t to infty}int_1^t 1/sqrt{x+4} dx#

by taking the antiderivative,

#=2lim_{t to infty}[sqrt{x+4}]_1^t#

#=2lim_{t to infty}(sqrt{t+4}-sqrt{5})=infty#

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Answer 2

Isaiah Allen

To use the Integral Test on the infinite series (\sum_{n=1}^\infty \frac{1}{\sqrt{n+4}}), follow these steps:

Formulate the corresponding integral: ( \int_{1}^{\infty} \frac{1}{\sqrt{n+4}} , dn).
Evaluate the integral.
Determine if the integral converges or diverges.
If the integral converges, then the series (\sum_{n=1}^\infty \frac{1}{\sqrt{n+4}}) converges. If the integral diverges, then the series also diverges.

So, for the given series, you'd integrate ( \frac{1}{\sqrt{n+4}} ) from 1 to infinity and analyze the result to determine whether the series converges or diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.