Question Video: Finding the Shape of a Curve Defined Parametrically Mathematics • Higher Education

We will investigate the convergence or divergence of the series βˆ‘_(𝑛 = 1)^(∞) 1/√(2𝑛 βˆ’ 1) using the integral test. Evaluate ∫_(1)^(∞) 1/√(2π‘₯ βˆ’ 1) dπ‘₯, if possible. Determine whether the series converges or diverges.

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Video Transcript

We will investigate the convergence or divergence of the series the sum from 𝑛 equals one to ∞ of one divided by the square root of two 𝑛 minus one using the integral test. Evaluate the integral from one to ∞ of one divided by the square root of two π‘₯ minus one with respect to π‘₯, if possible. Determine whether the series converges or diverges.

The first thing the question wants us to do is to evaluate the integral from one to ∞ of one divided by the square root of two π‘₯ minus one with respect to π‘₯. And in this case, we can see the top integrand is the composition of continuous functions. This means it will be continuous across its entire domain. And we can see the domain of our integrand is all values of π‘₯ strictly greater than one-half. So, our integrand is continuous across their entire interval of integration. However, we can see the upper limit of our integral is ∞. This means that our integral is improper, so we’re going to need to use our rules for improper integrals.

We recall from our rules of improper integrals. We can rewrite this integral as the limit as 𝑑 approaches ∞ of the integral from one to 𝑑 of one divided by the square root of two π‘₯ minus one with respect to π‘₯. And that’s, of course, only if this limit exists. Now, we can see the upper limit of our integral is approaching ∞. It’s never equal to ∞, and our integrand is continuous on the closed interval from one to 𝑑. This means we can apply any of our rules of integration to try and evaluate this integral.

To help us evaluate this integral, we’re going to use the 𝑒 substitution 𝑒 is equal to two π‘₯ minus one. Differentiating both sides of our substitution with respect to π‘₯, we get d𝑒 by dπ‘₯ is equal to two. And of course, we know d𝑒 by dπ‘₯ is not a fraction. However, when we’re integrating by using substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑒 is equal to two dπ‘₯. And in this case, it’s easier to divide both sides of our equation through by two. This gives us one-half d𝑒 is equal to dπ‘₯.

We’re almost ready to use our substitution. However, remember, since we’re using this on a definite integral, we need to change the limits of integration. The lower limit of our integral is when π‘₯ is equal to one. So, we substitute this into our equation for 𝑒. This gives us 𝑒 is equal to two times one minus one, which we can evaluate is equal to one. And we can do the same to find the new upper limit of our integral.

The upper limit of our integral was when π‘₯ was equal to 𝑑. We substitute this into our equation for 𝑒, and we get that 𝑒 is equal to two 𝑑 minus one. So, by using our 𝑒 substitution, the new upper limit of our integral is two 𝑑 minus one and the new lower limit of our integral is one.

Now, let’s look at our integrand. We know the square root of two π‘₯ minus one is equal to the square root of 𝑒. And we also know that dπ‘₯ is equal to one-half d𝑒. So, by using our 𝑒 substitution, we now have the limit as 𝑑 approaches ∞ of the integral from one to two 𝑑 minus one of one over the square root of 𝑒 multiplied by one-half with respect to 𝑒. And of course, we can simplify this by taking the factor of one-half outside of our integral. In fact, since this is a constant, we can just take it outside of our limit altogether.

We’re now ready to evaluate this integral. Remember, by using our laws of exponents, we can rewrite one over root 𝑒 as 𝑒 to the power of negative one-half. This means we can evaluate this integral by using the power rule for integration. We want to add one to our exponent of 𝑒 and then divide by this new exponent of 𝑒. So, we add one to our exponent of 𝑒, giving us 𝑒 to the power of one-half. And then, we divide by this new exponent of 𝑒. This gives us 𝑒 to the power of one-half divided by one-half.

And we can then simplify this expression. Instead of dividing by one-half, we’ll multiply by the reciprocal of one-half, which we know is two. And of course, we can simplify this even further. We can take the constant factor of two outside of our limit. This gives us the constant factor of two over two, which is, of course, equal to one. Next, we can use our laws of exponents to simplify 𝑒 to the power of one-half as the square root of 𝑒. So, we now have the limit as 𝑑 approaches ∞ of the square root of 𝑒 evaluated at the limits of integration, 𝑒 is equal to one and 𝑒 is equal to two 𝑑 minus one.

Now, we’ll evaluate this at the limit of integration. This gives us the limit as 𝑑 approaches ∞ of the square root of two 𝑑 minus one minus the square root of one. And of course, we know the square root of one is just equal to one. We’re now ready to evaluate our limit. We can see that our limit is as 𝑑 is approaching ∞. We can see that one remains constant as 𝑑 approaches ∞.

However, as 𝑑 approaches ∞, the square root of two 𝑑 minus one is growing and growing. In fact, it’s growing without bound. So, this term is approaching ∞. But then, this tells us our entire expression is growing without bound. So, our limit does not converge. And remember, the divergence or convergence of this limit is the same as the divergence or convergence of the integral we started with. So, by showing this limit did not converge. We’ve shown the integral from one to ∞ of one over the square root of two π‘₯ minus one with respect to π‘₯ does not converge.

The second part of this question wants us to use the integral test to determine the convergence or divergence of the sum from 𝑛 equals one to ∞ of one divided by the square root of two 𝑛 minus one. Since the question wants us to use the integral test, we’ll recall what the integral test tells us.

The integral test tells us if we have a function 𝑓 of π‘₯ which is continuous, positive, and decreasing for all values of π‘₯ greater than or equal to some constant π‘˜. And we have a sequence π‘Žπ‘› which is equal to 𝑓 evaluated at 𝑛 for all of our values of 𝑛. Then if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is convergent, then the sum from 𝑛 equals π‘˜ to ∞ of π‘Žπ‘› must also be convergent. However, if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is divergent, then our sum from 𝑛 equals π‘˜ to ∞ of π‘Žπ‘› must also be divergent.

Remember, we’ve already shown that the integral from one to ∞ of one over the square root of two π‘₯ minus one with respect to π‘₯ is divergent. And we can see this is very similar to the concluding statement of the second part of our integral test, where we set π‘˜ equals to one and our function 𝑓 of π‘₯ to be one divided by the square root of two π‘₯ minus one. So, to use the integral test, we’re going to need to show that our function 𝑓 of π‘₯ is continuous, positive, and decreasing for all values of π‘₯ greater than or equal to one.

Then, if we show all three of these prerequisites are true, then if we generate the sequence π‘Žπ‘› by taking 𝑓 evaluated at 𝑛 for each of our values of 𝑛, we would then be able to show by the integral test that the sum from 𝑛 equals one to ∞ of the sequence would be divergent. And if we input 𝑛 into our function 𝑓 of π‘₯, we can see this is exactly equal to the summand given to us in the question. So, if we show the prerequisites for the integral test are true, we can show that the series given to us in the question is divergent.

So, let’s start by showing that 𝑓 of π‘₯ is a continuous function for all values of π‘₯ greater than or equal to one. In fact, we’ve already shown that this is true. When we were using our rules for improper integrals, we argued that our integrand was continuous for all values of π‘₯ greater than one-half. And our integrand in this case was equal to 𝑓 of π‘₯. So, in particular, we know that our function 𝑓 of π‘₯ is continuous for all values of π‘₯ greater than or equal to one.

To see that our function 𝑓 of π‘₯ is positive, let’s take a closer look at our function 𝑓 of π‘₯. The numerator one is always positive. And the denominator is the positive square root of two π‘₯ minus one. Since we’re taking the positive square root, this is also positive. So, for all values of π‘₯, we’re taking the quotient of two positive numbers. This means 𝑓 of π‘₯ is positive for all values of π‘₯.

The last thing we need to do is show that 𝑓 of π‘₯ is a decreasing function. To do this, let’s consider what would happen if we were to increase our value of π‘₯. If we increase the value of π‘₯, then our entire denominator is getting larger. And remember, our denominator is always positive. So, now, we’re dividing by a larger positive number. And since our numerator of one is positive, whenever we increase the value of π‘₯, we’re dividing a positive number by a larger positive number. This means we’re making our number smaller. So, increasing the value of π‘₯ makes our function smaller. Therefore, our function 𝑓 of π‘₯ is decreasing.

So, we showed all of the prerequisites for the integral test are true. And since we’ve already shown the integral from one to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is divergent. This means the sum from 𝑛 equals one to ∞ of one divided by the square root of two 𝑛 minus one is also divergent.

Therefore, we were able to show the integral from one to ∞ of one divided by the square root of two π‘₯ minus one with respect to π‘₯ was divergent. And then, by using this information and the integral test, we were able to show the sum from 𝑛 equals one to ∞ of one divided by the square root of two 𝑛 minus one is also divergent.

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