### Video Transcript

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

We can use the integral test to
test this convergence. The integral test tells us that if
we have a function π of π₯ which is continuous, positive, and decreasing on the
interval between π and β and that π of π is equal to π π. Then, firstly, if the integral from
π to β of π of π₯ with respect to π₯ is convergent, so is the sum from π equals
π to β of π π. And secondly, if the integral from
π equals π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π
equals π to β of π π. Now, our series is the sum from π
equals one to β of the square root of the natural logarithm of π over π. And so, we can say that π π is
equal to the square root of the natural logarithm of π over π. Since π of π is equal to π π,
π 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 π of
π₯ is continuous, positive, and decreasing on the interval between one and β. Firstly, for continuity, we need
the inside of the square root, so thatβs 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 β. 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 itβs a positive square root. And the denominator is always
positive, since π₯ is greater than or equal to one. Therefore, weβve satisfied this
condition. Now, we need to check that our
function is decreasing between one and β. Now, itβs quite difficult to check
this without performing a calculation. So, we can find the values of π₯
for which π of π₯ is decreasing by differentiating π with respect to π₯. This will give us π prime of
π₯. Now, π of π₯ is a quotient. Therefore, we can use the quotient
rule. The quotient rule tells us that the
differential of π’ over π£ is equal to π£ multiplied by the differential of π’ minus
π’ multiplied by the differential of π£ all over π£ squared.

Now, in our case, π’ is equal to
the square root of the natural logarithm of π₯ and π£ is equal to π₯. Finding dπ£ 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 π’. And we can rewrite it as the
natural logarithm of π₯ to the power of one-half.

To find dπ’ 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, thatβs the differential of the
natural logarithm of π₯. And this is simply equal to one
over π₯. Now, dπ’ by dπ₯ and dπ£ by dπ₯ is
simply π’ prime and π£ prime. And these can be substituted into
the formula for the quotient rule. We obtain that π 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, weβre looking for the values
of π₯ such that π is decreasing. So, that means that π prime of π₯
is negative. Now, the denominator of π prime of
π₯, thatβs π₯ squared, is always positive since the square is always positive. Therefore, the values of π₯ for
which π 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 π 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 π equals two to β of the square root of the
natural logarithm of π over π. 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 π is
equal to two. Therefore, our value of π changes
to two. Now, we have that π of π₯ is
decreasing for π₯ is greater than 1.648. Therefore, weβve satisfied the last
condition to use the integral test and weβre able to use the integral test. So, letβs do that now.

We need to find out whether the
integral from two to β 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 π’ substitution. We can let π’ be equal to the
square root of the natural logarithm of π₯. Rearranging this, we have that π₯
is equal to π to the power of π’ squared. We can differentiate π₯ with
respect to π’ to find that dπ₯ by dπ’ is equal to two π’π to the π’ squared. An equivalent statement is dπ₯ is
equal to two π’π to the π’ squared dπ’. When π₯ is equal to two, π’ is
equal to the square root of the natural logarithm of two. And when π₯ is equal to β, π’ is
also equal to β. Weβre 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 β. Then, we have that the square root
of the natural algorithm of π₯ is π’. And π₯ is π to the π’ squared. Finally, we need to substitute in
dπ₯. Now, we have completed our π’
substitution. Weβre able to cancel out the π to
the π’ squared in the denominator with the other π to the π’ squared. And our integral becomes the
integral from the square root of the natural logarithm of two to β of two π’ squared
dπ’.

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