The nth Term Test for Divergence

The nth Term Test for Divergence: AP Calculus Guide

Time to dive into some serious math—pun intended! We're venturing into the nth Term Test for Divergence, part of the fantastic world of infinite sequences & series in AP Calculus BC. If you're in AP Calculus AB, feel free to surf the web elsewhere. For all you BC folks, fasten your seat belts, because we’re about to take off into the infinite! 🚀

Hold on to your calculators, folks! The nth Term Test for Divergence lets us know if a series is about to take a wild leap into infinity, aka diverge. The test boldly states:

If the limit of the nth term of a series as n approaches infinity is not zero, the series diverges. In math-speak: [ \text{if} \ \lim_{n \to \infty} a_n \neq 0, \ \sum a_n \ \text{diverges} ]

In simpler terms, if the individual terms of a series don't shrink to zero as you keep adding more and more of them, the series is off on an adventure to infinity and beyond. However, if they do shrink to zero, the series might still be playing coy—sometimes it converges and sometimes it doesn’t. The nth Term Test doesn't promise convergence; it just weeds out non-zero limiters from the crowd.

Let’s see how this test works with an example! Ready? Let’s go!

Imagine you’re given the series: [ \sum_{n=1}^{\infty} \frac{5n - 1}{n - 2} ]

1. Convert to Limit Notation ✏️: [ \lim_{n \to \infty} \frac{5n - 1}{n - 2} ]

2. Evaluate the Limit 📏: Simplify the limit expression by diving into infinity: [ \lim_{n \to \infty} \frac{n(5 - \frac{1}{n})}{n(1 - \frac{2}{n})} = \frac{5 - \frac{1}{\infty}}{1 - \frac{2}{\infty}} = \frac{5}{1} = 5 ] Don't sweat the small stuff like (\frac{1}{\infty}); it’s basically 0.

3. Conclude Your Conclusion 🤔: Since ( \lim_{n \to \infty} \frac{5n - 1}{n - 2} = 5 \neq 0 ), we can say: [ \sum_{n=1}^{\infty} \frac{5n - 1}{n - 2} \ \text{diverges} ]

And there you have it! The series goes off the rails into diverging territory.

Let’s see you strut your stuff. Try working out these sequences on your own:

1. [ \sum_{n=1}^{\infty} \frac{n^3 + 3n}{2n^3 - 5} ]

2. [ \sum_{n=1}^{\infty} \arctan(n) ]

Take them for a spin and see if they stay on track or veer into divergence city.

Solution 1:

1. Convert to Limit Notation ✏️: [ \lim_{n \to \infty} \frac{n^3 + 3n}{2n^3 - 5} ]

2. Evaluate the Limit 📏: Simplify the limit expression: [ \lim_{n \to \infty} \frac{n^3(1 + \frac{3}{n^2})}{n^3(2 - \frac{5}{n^3})} = \frac{1 + \frac{3}{\infty}}{2 - \frac{5}{\infty}} = \frac{1}{2} ]

3. Conclude Your Conclusion 🤔: Since ( \lim_{n \to \infty} \frac{n^3 + 3n}{2n^3 - 5} = \frac{1}{2} \neq 0 ), the series: [ \sum_{n=1}^{\infty} \frac{n^3 + 3n}{2n^3 - 5} \ \text{diverges} ]

Solution 2:

1. Convert to Limit Notation ✏️: Apply the same framework as before.

2. Evaluate the Limit 📏: Recall the behavior of ( \arctan(n) ) as ( n \to \infty ): [ \lim_{n \to \infty} \arctan(n) = \frac{\pi}{2} ]

3. Conclude Your Conclusion 🤔: Since ( \lim_{n \to \infty} \arctan(n) = \frac{\pi}{2} \neq 0 ), the series: [ \sum_{n=1}^{\infty} \arctan(n) \ \text{diverges} ]

To sum it up (pun intended), the nth Term Test for Divergence is a nifty tool in your math kit. If the limit of the nth term of a series doesn't zip off to zero, the series diverges into infinity. Remember, the nth Term Test doesn’t vouch for convergence; it simply gives divergence a thumbs up.

So go out there, tackle those series, and let the math geek within you thrive! Good luck, and may the ƒ(x) be with you. 📈✨