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Using the integral test, how do you show whether #1/(2n+3)# diverges or converges?

Answer 1

Cameron Alexander

#int_1^oo 1/(2x+3) dx = lim_(brarroo) int_1^b 1/(2x+3) dx#

#= lim_(brarroo) 1/2 int_1^b 1/(2x+3) 2dx#

#= lim_(brarroo) 1/2 ln(2x+3)|_1^b #

#= lim_(brarroo) [1/2 ln(2b+3) - 1/2 ln5]#

The integral, and hence the series, diverges.

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Answer 2

William Abdalla

To determine whether the series ∑(1/(2n + 3)) converges or diverges, we can use the integral test. First, we need to find the integral of the function f(x) = 1/(2x + 3) from 1 to infinity. Then, we compare the integral to the series.

The integral of f(x) from 1 to infinity is ∫[1,∞] (1/(2x + 3)) dx.

To integrate this function, we can use a substitution. Let u = 2x + 3, then du = 2 dx.

Now, the integral becomes ∫(1/u) * (1/2) du, integrated from 5 to infinity.

Integrating, we get: (1/2) * ln|u| evaluated from 5 to infinity.

Evaluating at the limits, we get: (1/2) * [ln(infinity) - ln(5)] = (1/2) * ∞ - (1/2) * ln(5) = ∞.

Since the integral diverges, by the integral test, the series ∑(1/(2n + 3)) also 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.