Integral Test — Does the Series Converge or Diverge?

Based on the supplied transcript. Punctuation and paragraph breaks have been added, with repeated fillers and false starts reduced for readability. Clearly identifiable mathematical expressions have been typeset. A complete mathematical explanation appears separately in the AI summary.

All right, we’re in the integral test section. There’s a specific set of questions where the author explicitly tells you to use the integral test. Here, the instructions don’t tell you which test to use.

That’s why I chose this question. I want to go back into the section and see which tests are available.

Here’s the integral test. Let me get a screenshot in case we need it. The question looks like an integral test problem, but we need to make sure it fits the theorem.

It could also be a p-series question. We aren’t doing sums or estimates here. We’ve reviewed the tests available in the section, and we go from there.

Is this a p-series, or should we use the integral test? You might look at it and think it doesn’t look like a p-series. Probably the integral test, but you never know what will happen. [Laughter.] That’s why you follow the theorems.

Let’s identify the terms. Let $a_n=f(n)$, with $f(n)$ equal to the expression given.

The theorem requires a continuous, positive, decreasing function. Is this function continuous, positive, and decreasing?

The interval starts at two and extends to infinity. That’s within the interval we’re considering.

Is it positive? Probably—but how do you know?

We’ve turned the expression into a function, so we can use differential calculus to determine whether it’s decreasing and positive.

The natural logarithm is positive for inputs greater than one. Let’s identify that: $\ln n>0$ for $n>1$.

Some people want to see the graph. I did that too. At one, the logarithm is zero. Below one, within its domain, it’s negative.

I don’t want you to depend on graphs because later courses won’t always have a visual connection. There’s mathematics used in physics; physics doesn’t describe mathematics. We’re talking about definitions and rules.

The real logarithm’s input can’t be zero or negative. You should know that at this point.

We know the expression is positive because the factors in the denominator are positive on the interval. That checks the positivity requirement.

Is it decreasing? Before taking derivatives, we could consider comparing consecutive terms or using induction.

At this level, showing a comparison might suffice, but you may not be allowed a calculator.

Is $1/(2\ln2)$ greater than the next term? You may not know that immediately without a calculator.

Instead of jumping to derivatives, we could compare the actual expressions. We want to know whether each term is greater than the next.

Sometimes you can rearrange that into a definitive answer; sometimes you can’t. Here, you end up comparing logarithms and a ratio involving $n$ and $n+1$.

That could lead to an induction argument. For me, let’s use differential calculus and take the derivative.

We’ll use the quotient rule, then the product rule in the denominator’s derivative. Let me put in parentheses and rearrange the pieces.

Do you see how much better typing is than writing? If you want to do all this with pencil and paper, be my guest. It takes longer.

I can’t write with a pencil anymore without my arm going numb. Even after many book signings, I switch to my left hand. I often use my left hand in the kitchen too. I’m ambidextrous now.

The derivative of the logarithm contributes $1/n$, and multiplying by $n$ gives one.

Now we’re checking whether the function increases or decreases. The denominator won’t be zero on the interval from two to infinity.

We can focus on the numerator’s sign. Setting it equal to zero gives $n=1/e$.

That’s less than one, so it isn’t in our interval. We can test a value in the interval to determine the derivative’s sign. It’s negative, so the function is decreasing.

We’ve checked another requirement. We also know the function is continuous on this interval from the logarithm’s domain and the form of the function.

We’ve checked continuity, positivity, and decreasing behavior. Now we can use the integral test.

If you don’t do this, you’ll lose credit. Don’t take it personally. [Laughter.]

Now integrate over the corresponding interval. Some students immediately do the integral and then wonder why they lose points despite getting the answer.

That’s what my professor told me: “You got the answers right, but you didn’t reference the theorems. How do I know you know what you’re doing? You could have looked over someone’s shoulder or guessed.”

He said he knew I wasn’t cheating, but he had to penalize me to make me wake up. I remember that.

Some students in that class had mental breakdowns. Now I deal with similar reactions online. Some listen and say, “I understand what you’re saying.”

Others say, “That’s ridiculous. I should be able to use AI and make $150,000 a year. This is America. Trump.” [Laughter.]

It looks like we aren’t getting $5,000, guys. It’s been a rough night, day, and year. I’m going to be homeless soon.

Now, how do we integrate this? Carefully.

I could write this using the logarithm directly—does that make sense? I’ll keep it simple.

Let $u=\ln x$. The lower limit becomes $\ln2$, and as $x$ tends to infinity, $u$ tends to infinity.

Now we have an integral from $\ln2$ to infinity of $1/u$ with respect to $u$.

Did you see what I did earlier with $\ln x$? It was the same substitution. You can recognize the change directly rather than write every substitution step.

These are things you’ll explore with me in the Math Monthly research program. We’re doing engineering this month, and I might extend it into December depending on my mouth.

I’m doing this to stay awake until eight. What do you do to stay awake? Hardcore math, baby.

The antiderivative is $\ln|u|$. Evaluating the improper integral gives infinity, so it diverges.

Finalize the solution. This is important: “Thus, by the integral test, the series is divergent.”

Let’s check whether I made a mistake. I make mistakes sometimes. This is 11.3.21. The book says divergent.

It is divergent, my children. That’s the integral test.

For ChatGPT preparing the description and thumbnail, the expression is the sum from $n=2$ to infinity of $1/(n\ln n)$. Please cite Stewart’s calculus book. Thank you.

Author Jonathan David applies the integral test to the series $\sum_{n=2}^{\infty}1/(n\ln n)$. The lesson emphasizes verifying the theorem’s hypotheses before integrating and explicitly connecting the integral’s behavior to the series.

Define the Corresponding Function

For $x\ge2$, both $x$ and $\ln x$ are positive. The denominator never vanishes, so $f$ is continuous and positive on $[2,\infty)$.

Verify Decreasing Behavior

Differentiating gives:

On $[2,\infty)$, the denominator is positive and $\ln x+1>0$. Therefore $f'(x)<0$, establishing that $f$ is decreasing throughout the required interval.

There is no need to find a critical number or test a numerical value once this sign is established.

Evaluate the Improper Integral

Use the substitution $u=\ln x$, with $du=dx/x$. Writing the improper integral as a limit:

The improper integral diverges. In particular, the lower-limit contribution is $\ln(\ln2)$, because the substitution changes the lower endpoint to $u=\ln2$.

State the Series Conclusion

The complete argument verifies continuity, positivity, and decreasing behavior, evaluates the improper integral, and applies the theorem. Exact presentation and grading requirements depend on the instructor.