The Alternating Series Test for Convergence - AP Calculus BC Study Guide
Introduction
Welcome to the mystical land of Calculus BC, where we learn to tame infinite sequences and series! This guide will equip you with the magic of the Alternating Series Test for Convergence. If you're navigating through Calc AB, you can teleport away now, but if Calc BC is your quest, read on! 🔮📚
Alternating Series Test Theorem
The Alternating Series Test (AST) is like the Basilisk of calculus—powerful and somewhat intimidating at first glance, but once you master it, you can do wonders. The theorem states that for an alternating series ∑(−1)ⁿ⋅aₙ (that's math code for summing up terms where the signs alternate), the series converges if:
- The limit of aₙ as n approaches infinity equals zero.
- The terms aₙ are decreasing.
In simple terms, if these two conditions are as true as a wizard's oath, your series will converge; otherwise, it diverges into the mathematical abyss. 🚀
Breaking Down the Theorem
To understand this theorem, let's embark on a journey through the land of the famous alternating harmonic series:
∑n=1∞(−1)ⁿ/n.
First, let's check if the limit of aₙ equals zero. Recall that aₙ is the main character here, while (−1)ⁿ is just the costume change for each term. So, in this series, aₙ = 1/n. Let's see if this term vanishes into thin air as n grows infinitely large.
Taking the limit of 1/n as n approaches infinity, we get: lim(n→∞) (1/n) = 0.
Voilà! The first criteria is satisfied. Now, to meet the second condition, we need to show that aₙ is getting smaller and smaller as n increases. In other words, each term in our series should be like the previous term's pocket-sized cousin.
If aₙ = 1/n, then aₙ+1 = 1/(n+1). Clearly: 1/n > 1/(n+1).
This shows that aₙ is indeed decreasing as n grows. Pop some random values of n in to reassure yourself. For example, n=2: 1/2 > 1/3.
Aha! Both our magical conditions are met. Therefore, the alternating harmonic series converges. ✨
Practice: Convergence or Divergence?
It's your turn to explore the mysterious realms of alternating series! For each of the following sequences, declare whether they converge or diverge.
- ∑n=1∞ ((−1)ⁿ+2⋅n⁵)/(n⁵+3)
- ∑n=2∞ (cos(nπ))/n
- ∑n=2∞ (−1)ⁿ⋅ln(n)
Solutions:
For the first series, our aₙ is n⁵/(n⁵+3). First, let's take the limit: lim(n→∞) (n⁵/(n⁵+3)) = 1 ≠ 0.
Since the limit doesn't tend to zero, this series joyously diverges.
Next up, the second series is a bit sneaky. Cos(nπ) oscillates like a complex dance move between -1 and 1, effectively meaning that cos(nπ) = (−1)ⁿ. Thus, our series looks like another harmonic sequence: ∑n=2∞ (−1)ⁿ/n.
We have already shown that this series meets the conditions for convergence. So ding ding ding, this series converges!
Finally, let’s check out the third series: aₙ = ln(n). Let's find the limit: lim(n→∞) ln(n) = ∞.
A limit that approaches infinity smells like divergence! Since the first condition isn’t met, we don’t need to check if aₙ is decreasing. This series is diverging faster than a wizard escaping a cursed forest.
Conclusion
Bravo! You've now mastered the art of the Alternating Series Test. With this powerful spell, you're ready to tackle convergence problems like a pro. Remember these two golden rules: the limit must approach zero, and the terms must decrease.
So, pack your wand, slide rule, and this guide as your spell book, and go forth to conquer Calculus BC! 📏🔮
