Video: Integrating Trigonometric Functions

Determine whether the series βˆ‘_(𝑛 = 1)^(∞) (√(ln 𝑛)/𝑛) converges or diverges.

06:52

Video Transcript

Determine whether the series, which is the sum from 𝑛 equals one to ∞ of the square root of the natural logarithm of 𝑛 over 𝑛, converges or diverges.

We can use the integral test to test this convergence. The integral test tells us that if we have a function 𝑓 of π‘₯ which is continuous, positive, and decreasing on the interval between π‘˜ and ∞ and that 𝑓 of 𝑛 is equal to π‘Ž 𝑛. Then, firstly, if the integral from π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is convergent, so is the sum from 𝑛 equals π‘˜ to ∞ of π‘Ž 𝑛. And secondly, if the integral from 𝑛 equals π‘˜ to ∞ of 𝑓 of π‘₯ with respect to π‘₯ is divergent, so is the sum from 𝑛 equals π‘˜ to ∞ of π‘Ž 𝑛. Now, our series is the sum from 𝑛 equals one to ∞ of the square root of the natural logarithm of 𝑛 over 𝑛. And so, we can say that π‘Ž 𝑛 is equal to the square root of the natural logarithm of 𝑛 over 𝑛. Since 𝑓 of 𝑛 is equal to π‘Ž 𝑛, 𝑓 of π‘₯ is equal to the square root of the natural logarithm of π‘₯ over π‘₯.

We also have that π‘˜ is equal to one. We now need to check whether 𝑓 of π‘₯ is continuous, positive, and decreasing on the interval between one and ∞. Firstly, for continuity, we need the inside of the square root, so that’s the natural logarithm of π‘₯, to be nonnegative. If the natural logarithm of π‘₯ is greater than or equal to zero, this means that π‘₯ must be greater than or equal to one. Since our value of π‘˜ is one, we are looking at the interval between one and ∞. Therefore, π‘₯ is always greater than or equal to one. And so, the numerator of our function is continuous. The only other discontinuity may occur when the denominator of our fraction is equal to zero. However, π‘₯ is always greater than or equal to one, and so π‘₯ cannot be equal to zero. Therefore, our function is continuous on the interval.

Next, we need to check whether our function is positive over this interval. The square root function in the numerator is always positive, since it’s a positive square root. And the denominator is always positive, since π‘₯ is greater than or equal to one. Therefore, we’ve satisfied this condition. Now, we need to check that our function is decreasing between one and ∞. Now, it’s quite difficult to check this without performing a calculation. So, we can find the values of π‘₯ for which 𝑓 of π‘₯ is decreasing by differentiating 𝑓 with respect to π‘₯. This will give us 𝑓 prime of π‘₯. Now, 𝑓 of π‘₯ is a quotient. Therefore, we can use the quotient rule. The quotient rule tells us that the differential of 𝑒 over 𝑣 is equal to 𝑣 multiplied by the differential of 𝑒 minus 𝑒 multiplied by the differential of 𝑣 all over 𝑣 squared.

Now, in our case, 𝑒 is equal to the square root of the natural logarithm of π‘₯ and 𝑣 is equal to π‘₯. Finding d𝑣 by dπ‘₯ is fairly straightforward. We differentiate π‘₯ and simply get one. Now, when we differentiate the square root of the natural logarithm of π‘₯, we need to use the chain rule. Now, we may find it easier to rewrite 𝑒. And we can rewrite it as the natural logarithm of π‘₯ to the power of one-half.

To find d𝑒 by dπ‘₯, we multiply by the power, decrease the power by one, and finally multiply by the differential of the inside of the function. So, that’s the differential of the natural logarithm of π‘₯. And this is simply equal to one over π‘₯. Now, d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯ is simply 𝑒 prime and 𝑣 prime. And these can be substituted into the formula for the quotient rule. We obtain that 𝑓 prime of π‘₯ is equal to π‘₯ multiplied by one-half multiplied by one over π‘₯ times the natural logarithm of π‘₯ to the power of negative one-half minus the square root of the natural logarithm of π‘₯ all over π‘₯ squared.

Now, we’re looking for the values of π‘₯ such that 𝑓 is decreasing. So, that means that 𝑓 prime of π‘₯ is negative. Now, the denominator of 𝑓 prime of π‘₯, that’s π‘₯ squared, is always positive since the square is always positive. Therefore, the values of π‘₯ for which 𝑓 prime is negative occur when the numerator of the fraction is negative. So, this is when the square root of the natural logarithm of π‘₯ is greater than π‘₯ multiplied by one-half multiplied by one over π‘₯ multiplied by the natural logarithm of π‘₯ to the power of negative one-half. Here, we can cancel out the one over π‘₯ and the π‘₯.

We can also rewrite the natural logarithm of π‘₯ to the power of negative one-half as one over the square root of the natural logarithm of π‘₯. Therefore, the right-hand side of our inequality becomes one over two times the square root of the natural logarithm of π‘₯. We can multiply both sides by the square root of the natural logarithm of π‘₯. We obtain that 𝑓 of π‘₯ is decreasing for values of π‘₯, such that the natural logarithm of π‘₯ is greater than one-half. Using a calculator, we find that this is values of π‘₯ which are greater than 1.648 and so on. Therefore, we found that our function is not decreasing over the entirety of our interval.

However, we are able to rewrite our series. We can take the first term out of our series so that our series is equal to the square root of the natural logarithm of one over one plus the sum from 𝑛 equals two to ∞ of the square root of the natural logarithm of 𝑛 over 𝑛. Since the natural logarithm of one is equal to zero, this term on its own is simply zero. And so, this new sum that we found is in fact equal to the original sum. However, we are summing from 𝑛 is equal to two. Therefore, our value of π‘˜ changes to two. Now, we have that 𝑓 of π‘₯ is decreasing for π‘₯ is greater than 1.648. Therefore, we’ve satisfied the last condition to use the integral test and we’re able to use the integral test. So, let’s do that now.

We need to find out whether the integral from two to ∞ of the square root of the natural logarithm of π‘₯ over π‘₯ with respect to π‘₯ is convergent or divergent. In order to solve this integral, we can in fact use a 𝑒 substitution. We can let 𝑒 be equal to the square root of the natural logarithm of π‘₯. Rearranging this, we have that π‘₯ is equal to 𝑒 to the power of 𝑒 squared. We can differentiate π‘₯ with respect to 𝑒 to find that dπ‘₯ by d𝑒 is equal to two 𝑒𝑒 to the 𝑒 squared. An equivalent statement is dπ‘₯ is equal to two 𝑒𝑒 to the 𝑒 squared d𝑒. When π‘₯ is equal to two, 𝑒 is equal to the square root of the natural logarithm of two. And when π‘₯ is equal to ∞, 𝑒 is also equal to ∞. We’re now ready to perform our substitution.

We can start by substituting in the bounds of our integral, which will be the square root of the natural logarithm of two and ∞. Then, we have that the square root of the natural algorithm of π‘₯ is 𝑒. And π‘₯ is 𝑒 to the 𝑒 squared. Finally, we need to substitute in dπ‘₯. Now, we have completed our 𝑒 substitution. We’re able to cancel out the 𝑒 to the 𝑒 squared in the denominator with the other 𝑒 to the 𝑒 squared. And our integral becomes the integral from the square root of the natural logarithm of two to ∞ of two 𝑒 squared d𝑒.

We can integrate this by raising the power of 𝑒 by one and dividing by the new power. So, now we have two-thirds 𝑒 cubed between our two bounds. Substituting in our bounds, we obtain this. And even though the second term here is constant, the limit as 𝑒 tends to ∞ of two-thirds 𝑒 cubed is in fact infinite. Therefore, we have that this integral is divergent. So, by the integral test, our series from 𝑛 equals two to ∞ diverges. Since this series is equal to the one in the question, we have found our solution, which is that our series is divergent.

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