The Integral Test for Convergence: Your Ultimate AP Calculus BC Study Guide
Introduction
Welcome to the enchanting world of AP Calculus BC, where integrals are celebrated and sequences never seem to end! 🧮 Hold on to your graphing calculators because today, we’re diving headfirst into the Integral Test for Convergence. Trust me, by the end, you'll be testing infinite series like a pro, even if your head occasionally feels like it's going to explode with infinite wisdom!
The Integral Test Theorem
First things first, let's introduce our star player: the Integral Test Theorem. 🏆 Imagine it as the calculus equivalent of a mystical oracle that tells us whether our infinite series will converge (stay finite) or diverge (run away to infinity).
Here’s the official sounding bit of magic: If you have a positive, decreasing function ( f(x) ) over the interval ([k, \infty)) with a corresponding sequence ( a_n = f(n) ), then:
 If the integral (\int_k^{\infty} f(x) , dx) converges to a finite number, then the series (\sum_{n=k}^{\infty} a_n) also converges.
 If the integral (\int_k^{\infty} f(x) , dx) diverges (heads to infinity), then the series (\sum_{n=k}^{\infty} a_n) also diverges.
Got that? Good! Now let's explore how to use this test without feeling like we need a wizard's hat. 🧙♂️
Breaking Down the Integral Test Theorem
First up, our function needs to be positive and decreasing. No, this isn't a vague dating profile requirement; it's a serious mathematical condition! A positive function never dips below zero, and a decreasing function means it gets smaller or stays the same as ( x ) increases.
Picture the function ( f(x) = \frac{1}{x \ln(x)} ) over the interval ([2, \infty) ). This function fits the bill nicely. It’s kind of like the mathematical version of a perfectly ripe avocado – it's rare but oh so satisfying when you find it! 🥑
Now, dress ( f(x) ) up in its discrete sequence form: [ \sum_{n=2}^{\infty} \frac{1}{n \ln(n)}. ]
Pretty snazzy, right? But wait, there’s more! Let's apply the Integral Test to verify if our series is a keeper or a dumpster fire of divergence.
Applying the Integral Test
Time to roll up our calculus sleeves and tackle this integral. We'll need ( u )substitution to solve:

Substitute ( u = \ln(x) ) and find ( du = \frac{1}{x} , dx ).

The integral becomes: [ \int \frac{1}{u} , du = \ln u + C. ]

Backsubstitute ( u = \ln(x) ): [ \int_2^{\infty} \frac{1}{x \ln(x)} , dx = \ln \ln(x) \bigg_2^{\infty}. ]
Evaluating this integral over the bounds results in: [ \ln \ln(\infty)  \ln \ln(2) = \infty. ]
Houston, we have divergence! This integral evaluates to infinity, so according to our thematic oracle, the series (\sum_{n=2}^{\infty} \frac{1}{n \ln(n)} ) also diverges. 🔮✨
Integral Test Theorem Practice
Ready for some handson fun? Grab your practice problems below and see if you can determine the fate of each series.

State the integral test and its conditions. Remember, don't just copypaste; understand!

Determine if the series converges or diverges: [ \sum_{n=0}^{\infty} \frac{1}{1+n^2} ]

Determine if the series converges or diverges: [ \sum_{n=0}^{\infty} \frac{n^3}{n^4+1} ]
Solutions
Integral Test Conditions: You need a function that’s positive, decreasing over an interval, and matches your series.

Series ( \sum_{n=0}^{\infty} \frac{1}{1+n^2} ): The integral (\int_0^{\infty} \frac{1}{1+x^2} , dx) is a classic arctangent integral, yielding (\frac{\pi}{2}), a finite result. Therefore, the series converges. 🎉

Series ( \sum_{n=0}^{\infty} \frac{n^3}{n^4+1} ): Using ( u = x^4 + 1 ) and ( du = 4x^3 , dx ), the integral transforms, yielding (\frac{\ln(u)}{4}), and evaluating this across the bounds results in infinity. Thus, the series diverges. 🚫
Conclusion
Congratulations, you've journeyed through the fantastical land of infinite series and emerged victorious with the power of the Integral Test! Remember, practice makes perfect. Now, go forth and conquer more integrals and series like the calculus wizard you are! 🧙♂️💫