Video Transcript
Use the comparison test to determine whether the sum from 𝑛 equals two to ∞ of one divided by the natural logarithm of 𝑛 is convergent or divergent.
The question asks us to use the comparison test to determine whether the series is convergent or divergent. Since we don’t know whether the series is convergent or divergent, it can be tricky to decide which series to compare our series to. By using what we know about 𝑃-series, we know that the sum from 𝑛 equals two to ∞ of one divided by 𝑛 to the 𝑃th power will converge when 𝑃 is greater than one and diverge if 𝑃 is less than or equal to one. This gives us multiple series, which we know the convergence and divergence of, which we can use to compare the series in our question to.
Since we want to use the comparison test, let’s compare the first term in our sum of one divided by the natural log of 𝑛 and the first term of the sum of one divided by 𝑛 to the 𝑃th power. The first term in the sum from 𝑛 equals two to ∞ of one divided by the natural log of 𝑛 is just one divided by the natural log of two. And similarly, the first term in the sum from 𝑛 equals two to ∞ of one divided by 𝑛 to the 𝑃th power is one divided by two to the power of 𝑃. If we set these values to be equal and we solve for 𝑃, we will find the value of 𝑃, which is the tipping point between being a lower bound and an upper bound for our first term.
If we take reciprocals of both sides of our equation, we get the natural log of two is equal to two to the 𝑃th power. We can then take the log base two of both sides of our equation. And we know that the log base two of two to the power of 𝑃 is just equal to 𝑃. And if we evaluate the log base two of the natural log of two, we get an answer which is approximately equal to negative 0.53. So, what we have shown is if we choose 𝑃 to be greater than negative 0.53, then the first term in our sum of the reciprocal of the natural logarithm function is bigger than the first term in the sum of one divided by 𝑛 to the power of 𝑃.
This is a useful result since we know the 𝑃-series is divergent when 𝑃 is less than or equal to one. And we’ve shown that the first term of our series will be bounded below by our 𝑃-series. Since we can bound our first term below by the 𝑃-series when 𝑃 is greater than negative 0.53 and we know that the 𝑃-series diverges if 𝑃 is less than or equal to one. Let’s try setting 𝑃 equal to one and using the comparison test with the harmonic series.
One version of the comparison test tells us that if the sum from 𝑛 equals two to ∞ of the sequence 𝑎 𝑛 diverges and zero is less than or equal to 𝑎 𝑛 less than equal to 𝑏 𝑛 when 𝑛 is greater than or equal to two. Then we can conclude that the sum from 𝑛 equals two to ∞ of the sequence 𝑏 𝑛 must also diverge. We want to use the comparison test with the harmonic series and our series in our question, the sum from 𝑛 equals two to ∞ of the reciprocal of the natural logarithm function.
What this gives us is that if the harmonic series from 𝑛 equals two to ∞ is divergent and zero is less than or equal to one divided by 𝑛 is less than or equal to the reciprocal of the natural logarithm function when 𝑛 is greater than or equal to two. Then we could conclude by using the comparison test that the sum from 𝑛 equals two to ∞ of the reciprocal of the natural logarithm function must also diverge.
We know that a 𝑃-series will be divergent if 𝑃 is less than or equal to one. So, the sum from 𝑛 equals two to ∞ of one divided by 𝑛 must be divergent. This means all we need to show is that zero is less than or equal to one over 𝑛 is less than or equal to the reciprocal of the natural logarithm function when 𝑛 is greater than or equal to two. First, we know it’s that when 𝑛 is greater than or equal to two, then the natural logarithm of 𝑛 must be positive. Now, we know it’s that if we have two positive numbers 𝑎 and 𝑏, then one divided by 𝑎 being less than one divided by 𝑏 is equivalent to saying that 𝑎 is bigger than 𝑏.
In our case, we have two positive denominators, 𝑛 and the natural logarithm of 𝑛. So, we can just compare the size of the denominators. To determine which is larger, we’re going to consider the difference between the two functions. So, we set the function 𝑓 of 𝑥 to be equal to 𝑥 minus the natural logarithm of 𝑥, where we remember that 𝑥 must be greater than or equal to two. We notice that our function evaluated at two is equal to two minus the natural logarithm of two, which is approximately 1.3, which is positive. So, when 𝑥 is equal to two, 𝑥 is greater than the natural logarithm of 𝑥.
If we then take the derivative of our function 𝑓 of 𝑥, we can see the rate of change of this difference. We see that our derivative function 𝑓 prime of 𝑥 is equal to the derivative of 𝑥, which is just one, minus the derivative of the natural logarithm of 𝑥, which is one over 𝑥. Now, we know it’s that since 𝑥 is greater than or equal to two, our derivative function 𝑓 prime of 𝑥 is strictly greater than zero. Since 𝑓 prime of 𝑥 is positive, we can conclude that our function 𝑓 of 𝑥 must be an increasing function. This means that the difference between 𝑥 and the natural logarithm of 𝑥 is getting bigger and bigger. So, we know that 𝑥 is greater than the natural logarithm of 𝑥 when 𝑥 is greater than or equal to two.
Now, since both 𝑥 and the natural logarithm of 𝑥 are positive and we’ve just shown that 𝑥 is greater than the natural logarithm of 𝑥 when 𝑥 is greater than or equal to two. We can conclude that the reciprocal one over 𝑥 is less than the reciprocal of the natural logarithm of 𝑥 when 𝑥 is greater than or equal to two. Therefore, since this must also be true if we change the 𝑥 for an 𝑛, we have shown that zero is less than or equal to one over 𝑛 is less than or equal to one over the natural logarithm of 𝑛 when 𝑛 is greater than or equal to two.
Therefore, we have shown both of our prerequisites for using the comparison test with the harmonic series are true. And therefore, we can conclude by using the comparison test with the harmonic series the sum from 𝑛 equals two to ∞ of the reciprocal of the natural logarithm function diverges.
