Question Video: Using the Integral Test on the Harmonic Series | Nagwa Question Video: Using the Integral Test on the Harmonic Series | Nagwa

Question Video: Using the Integral Test on the Harmonic Series Mathematics • Higher Education

Use the integral test to determine whether the series β_(π = 1)^(β) 1/n is convergent or divergent.

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Video Transcript

Use the integral test to determine whether the series the sum from π equals one to β of one divided by π is convergent or divergent.

The question is asking us to use the integral test to determine the convergence or divergence of this series. Weβll recall the integral test tells us if we have a function π of π₯, which is continuous, positive, and decreasing for values of π₯ greater than or equal to one and π evaluated at π is equal to sum sequence π π. Then we can conclude, if the integral from one to β of π of π₯ with respect to π₯ converges, then our series the sum from π equals one to β of π π also converges. However, if the integral from one to β of π of π₯ with respect to π₯ diverges, then our series the sum from π equals one to β of π π also diverges.

The integral test gives us a way of turning a question about the convergence or divergence of a series into a question about the convergence or divergence of an integral. We want to use the integral test on the series the sum from π equals one to β of one divided by π. The summand of our series is one divided by π. So, weβll set π of π₯ equal to one divided by π₯. We then need to show that π of π₯ is continuous, positive, and decreasing for all values of π₯ greater than or equal to one.

We know that one is continuous and π₯ is continuous. So, one divided by π₯ is the quotient of two continuous functions. And the quotient of two continuous functions is always continuous, unless the denominator is equal to zero. However, since weβre only interested in values of π₯ greater than or equal to one, the denominator of π₯ is never equal to zero. So our function π of π₯ is equal to one divided by π₯ is continuous on this interval. When π₯ is greater than or equal to one, in particular, π₯ is positive and π of π₯ is one divided by π₯. And one divided by a positive number is positive. So, our function π of π₯ is positive on this interval.

Now, we need to show that our function π of π₯ is decreasing. Thereβs a few different ways of showing this. For example, we could calculate π prime of π₯ by using the power rule for differentiation. This gives us negative one divided by π₯ squared. π₯ squared is always positive. So, one divided by π₯ squared is positive. And then, we multiply this by negative one. So, this is always negative; itβs less than zero. And π prime of π₯ represents the slope of our function. So, when π₯ is greater than or equal to one, the slope of our function is negative. This means itβs decreasing.

So, weβve shown our function π of π₯ is continuous, positive, and decreasing for values of π₯ greater than or equal to one. And we chose our function π of π₯ so that π evaluated at π is one divided by π. This means weβve turned our question about the convergence or divergence of our series into one about the convergence or divergence of an integral. We want to determine the convergence or divergence of the integral from one to β of one divided by π₯ with respect to π₯.

Then by the integral test, this will be the same as the convergence or divergence of our series. We determine the convergence or divergence of our integral by determining the convergence or divergence of the limit as π‘ approaches β of the integral from one to π‘ of one divided by π₯ with respect to π₯. And we can calculate the integral of one divided by π₯ with respect to π₯. Itβs equal to the natural logarithm of the absolute value of π₯ plus a constant of integration π. And of course, when we were showing that we could use the integral test, we already showed that our function was continuous on this interval.

So, this gives us the limit as π‘ approaches β of the natural logarithm of the absolute value of π₯ evaluated at the limits of our integral π₯ is equal to one and π₯ is equal to π‘. Evaluating this at the limits of our integral gives us the limit as π‘ approaches β of the natural logarithm of the absolute value of π‘ minus the natural logarithm of the absolute value of one. One is a positive number. So, the absolute value of one is just equal to one.

Then, using our laws of logarithms, the natural logarithm of one is equal to zero. And π‘ is the upper limit of our integral, and the lower limit of our integral is one. So, π‘ is greater than or equal to one. This means the absolute value of π‘ is just equal to π‘. This gives us the limit as π‘ approaches β of the natural logarithm of π‘. And we know this limit is equal to positive β. However, when a limit is equal to positive or negative β, we still say that it diverges. This means weβve shown that our integral was divergent. This means we can use the integral test to conclude that our series must also be divergent.

Therefore, weβve shown by using the integral test, the sum from π equals one to β of one divided by π is divergent.

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