### Defining Convergent and Divergent Infinite Series: AP Calculus BC Study Guide

#### Introduction

Welcome to the grand finale of AP Calculus BC, Unit 10! 🎉 We’re delving into the realm of infinite sequences and series. This unit explores whether a series can settle on a sum as the number of terms shoots up to infinity. It's like seeing if an endless buffet eventually stops serving dishes (spoiler alert: sometimes it does, sometimes it doesn't). 🥢

#### Sequences: A Prelude to Series

Hold your mathematical horses; before we dive into series, we need a pitstop at sequences. A sequence is a list of numbers that follow a specific rule or pattern. Imagine it as a never-ending line of dominos, each one following the other in a precise order. 📏

Here’s how we represent sequences: [ {a_n}_{n=1}^\infty ]

#### Terms in a Sequence

Let’s go ahead and dig into the terms of some sequences with a sprinkle of examples.

**Example 1:**
[ \left{ \frac{1}{n} \right}_{n=1}^\infty ]
Plugging in values of ( n ), the sequence looks like:
[ {1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n}, \frac{1}{n+1}, \ldots } ]

This is known as the **harmonic sequence**. It's like a musical scale—but for math enthusiasts only. 🎵

**Example 2:**
[ \left{ \frac{(-1)^n \cdot n!}{2^n} \right}_{n=1}^\infty ]
This one takes a bit more algebra muscle. Let’s flex those biceps: 💪
[ a_1 = \frac{(-1)^1 \cdot 1!}{2^1} = -\frac{1}{2} ]
[ a_2 = \frac{(-1)^2 \cdot 2!}{2^2} = \frac{1}{2} ]
[ a_n = \frac{(-1)^n \cdot n!}{2^n} ]

Since this sequence alternates between positive and negative values (just like a seesaw 🌊), we call it an **alternating sequence**. You're getting the hang of it!

#### Limits of Sequences

Just like functions, sequences have limits too! Instead of looking at the limit as ( x ) approaches a number, here we look at the limit as ( n ) approaches ∞ (because, why not? Infinity is more fun!). [ \lim_{n \to \infty} a_n ]

**Convergent Sequence:**A sequence where the limit exists and is finite.**Divergent Sequence:**A sequence where the limit doesn't exist or is infinite.

#### Convergence and Divergence of Sequences

Let’s roll up our sleeves and determine whether the following sequences converge or diverge.

**Example 1:**
[ \left{ \frac{(-1)^n}{n} \right}*{n=1}^\infty ]
[ \lim*{n \to \infty} \frac{(-1)^n}{n} = 0 ]
Since the limit is finite, this sequence converges. 🎯

**Example 2:**
[ \left{ \frac{n^2 + 1}{n} \right}*{n=1}^\infty ]
[ \lim*{n \to \infty} \frac{n^2 + 1}{n} = \infty ]
The result is infinite, hence this sequence diverges. 🚀

**Example 3:**
[ \left{ (-1)^n \right}_{n=1}^\infty ]
This sequence flips between 1 and -1 forever (it's got commitment issues). So, the limit does not exist, thus the sequence diverges.

#### Series: Summing It Up

Alright, now let’s graduate to series, which is just a fancy name for the sum of the terms of a sequence. The notation for a series looks like this:
[ s_n = \sum_{i=1}^n a_i ]
Where ( s_n ) is the **nth partial sum**.

#### Infinite Series and Partial Sums

For an infinite series: [ s_\infty = \lim_{n \to \infty} \sum_{i=1}^n a_i ]

Let's tackle an example to show this in action: Given: ( a_n = \frac{1}{n^2 + n} ), [ a_n = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} ]

By telescoping (not like with stars, but with math 📡): [ s_n = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) ] Each term cancels the next (like a magic trick 🪄): [ s_n = 1 - \frac{1}{n+1} ] So, [ \lim_{n \to \infty} s_n = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 ] This infinite series converges! Huzzah!

#### Convergence and Divergence of Series

Much like sequences, series can also converge or diverge:

**Convergent Series:**The infinite sum ( s_\infty ) exists and is finite.**Divergent Series:**The infinite sum ( s_\infty ) doesn't exist or is infinite.

#### Series Properties: Handy Tools

Got your toolbox ready? Here’s what you’ll need:

- If ( s_n ) and ( t_n ) are convergent, so is ( c \cdot s_n ) for any constant ( c ).
- Adding or subtracting convergent series (( s_n \pm t_n )) also gives a convergent series.

### Practice Problems

[ a_n = \frac{1}{e^n} ] [ a_{n+1} = e^{-(n+1)} ] [ s_1 = e^{-1} ] [ s_2 = e^{-1} + e^{-2} ] [ s_3 = e^{-1} + e^{-2} + e^{-3} ]

[ a_n = \frac{n}{(n+2)!} ] [ a_{n+1} = \frac{(n+1)}{(n+3)!} ] [ s_1 = \frac{1}{6} ] [ s_2 = \frac{1}{4} ] [ s_3 = \frac{11}{40} ]

### Conclusion

You're now armed with the knowledge to conquer infinite sequences and series. Keep practicing, and soon, you'll be a calculating wizard! 🧙 Continue your journey to mathematical greatness in unit 10! 🚀

Now go forth and ace your AP Calculus BC exam—this guide is your Excalibur! ⚔️