Question: How can we add infinitely many numbers?

Answer: Use partial sums!

Partial Sums

The nth partial sum $S_n$ is the sum of the first $n$ terms:

$S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n$

Examples:

• $S_1 = a_1$
• $S_2 = a_1 + a_2$
• $S_3 = a_1 + a_2 + a_3$

Convergence of a Series

The series $\sum_{n=1}^{\infty} a_n$ converges to $S$ if:

$\lim_{n \to \infty} S_n = S$

where $S_n$ is the sequence of partial sums.

If the limit exists (and is finite), the series converges.

If the limit doesn't exist or is infinite, the series diverges.

💡 Key Idea: A series converges if its partial sums approach a finite number!

Example 1: Finite Geometric Series

Sum: $1 + 2 + 4 + 8 + \cdots + 2^{n-1}$

Formula: For geometric series with first term $a$ and ratio $r$:

$S_n = a\frac{1 - r^n}{1 - r} \quad (r \neq 1)$

Here: $a = 1$, $r = 2$

$S_n = \frac{1 - 2^n}{1 - 2} = \frac{1 - 2^n}{-1} = 2^n - 1$

Check: $S_1 = 1$, $S_2 = 3$, $S_3 = 7$, $S_4 = 15$ ✓

Infinite Geometric Series

$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \cdots$

Partial sum: $S_n = a\frac{1 - r^n}{1 - r}$ (if $r \neq 1$)

Take limit:

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1 - r^n}{1 - r}$

Case 1: If $|r| < 1$, then $r^n \to 0$ as $n \to \infty$

$S = \frac{a}{1 - r}$

Series converges!

Case 2: If $|r| \geq 1$, then $r^n$ doesn't approach 0

Series diverges!

Geometric Series Formula

$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad \text{if } |r| < 1$

Diverges if $|r| \geq 1$

🎯 MEMORIZE THIS! Most important series formula!

Example 2: Evaluate $\sum_{n=0}^{\infty} \frac{1}{2^n}$

This is geometric with $a = 1$ and $r = \frac{1}{2}$.

Since $|r| = \frac{1}{2} < 1$, it converges:

$S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$

Check: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 2$ ✓

Example 3: Evaluate $\sum_{n=1}^{\infty} \frac{3}{5^n}$

Rewrite with $n$ starting at 0:

$\sum_{n=1}^{\infty} \frac{3}{5^n} = 3\sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n$

$= 3\left[\sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1\right]$

$= 3\left[\frac{1}{1-\frac{1}{5}} - 1\right]$

$= 3\left[\frac{5}{4} - 1\right] = 3 \cdot \frac{1}{4} = \frac{3}{4}$

Or directly: First term is $a = \frac{3}{5}$, ratio $r = \frac{1}{5}$

$S = \frac{\frac{3}{5}}{1 - \frac{1}{5}} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$

Telescoping Series

A series where most terms cancel out!

General form: $\sum_{n=1}^{\infty} (b_n - b_{n+1})$

Partial sum: $S_n = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \cdots + (b_n - b_{n+1})$

Most terms cancel: $S_n = b_1 - b_{n+1}$

Limit: $S = \lim_{n \to \infty} S_n = b_1 - \lim_{n \to \infty} b_{n+1}$

If $\lim_{n \to \infty} b_{n+1}$ exists, the series converges!

Example 4: Telescoping Series

Evaluate $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.

Step 1: Use partial fractions

$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

$1 = A(n+1) + Bn$

Set $n = 0$: $1 = A$

Set $n = -1$: $1 = -B$, so $B = -1$

$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

Step 2: Write out partial sum

$S_n = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right)$

$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$

Step 3: Most terms cancel (telescope!)

$S_n = 1 - \frac{1}{n+1}$

Step 4: Take limit

$S = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1 - 0 = 1$

Answer: $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$

The Harmonic Series

$\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$

Question: Does it converge?

Intuition: Terms approach 0, so maybe it converges?

Answer: IT DIVERGES! 🤯

Proof (grouping argument):

$S = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots$

$> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots$

$= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = \infty$

The harmonic series diverges!

⚠️ Important: Just because $a_n \to 0$ doesn't mean $\sum a_n$ converges!

The nth Term Test for Divergence

If $\lim_{n \to \infty} a_n \neq 0$, then $\sum_{n=1}^{\infty} a_n$ diverges.

Contrapositive: If $\sum a_n$ converges, then $\lim a_n = 0$.

⚠️ WARNING: The converse is FALSE! $\lim a_n = 0$ does NOT imply convergence (harmonic series is a counterexample).

Example 5: Use nth Term Test

Does $\sum_{n=1}^{\infty} \frac{n}{2n+1}$ converge?

Check limit:

$\lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2} \neq 0$

By nth Term Test, the series diverges.

Properties of Convergent Series

If $\sum a_n = A$ and $\sum b_n = B$ (both converge), then:

1. Sum: $\sum (a_n + b_n) = A + B$

2. Constant Multiple: $\sum c \cdot a_n = c \cdot A$

3. Difference: $\sum (a_n - b_n) = A - B$

Note: Can't multiply series directly! $(\sum a_n)(\sum b_n) \neq \sum a_n b_n$

Example 6: Use Properties

If $\sum_{n=1}^{\infty} a_n = 5$ and $\sum_{n=1}^{\infty} b_n = 3$, find:

$\sum_{n=1}^{\infty} (2a_n - 3b_n)$

Solution:

$\sum (2a_n - 3b_n) = 2\sum a_n - 3\sum b_n = 2(5) - 3(3) = 10 - 9 = 1$

Changing Index

You can shift the starting index of a series (but it changes the value for geometric series!).

Example: $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ (if $|r| < 1$)

But: $\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}$ (missing the first term!)

Always check where the sum starts!

⚠️ Common Mistakes

Mistake 1: Thinking $\lim a_n = 0$ Implies Convergence

WRONG: "Since $\lim \frac{1}{n} = 0$, the series $\sum \frac{1}{n}$ converges"

RIGHT: Harmonic series diverges even though terms approach 0!

$\lim a_n = 0$ is necessary but not sufficient for convergence.

Mistake 2: Wrong Geometric Series Formula

For $\sum_{n=1}^{\infty} ar^n$ (starting at $n=1$):

First term is $ar$ (not $a$), so sum is $\frac{ar}{1-r}$

Or rewrite: $\sum_{n=1}^{\infty} ar^n = ar\sum_{n=0}^{\infty} r^n = ar \cdot \frac{1}{1-r} = \frac{ar}{1-r}$

Mistake 3: Using nth Term Test to Prove Convergence

nth Term Test can only prove DIVERGENCE!

If $\lim a_n = 0$, you learn nothing from this test.

Mistake 4: Arithmetic with Divergent Series

Can't add/subtract/multiply divergent series using the properties!

Properties only work when BOTH series converge.

📝 Practice Strategy

1. Geometric series: Identify $a$ and $r$, check $|r| < 1$, use formula $\frac{a}{1-r}$
2. Telescoping: Use partial fractions, write out terms to see cancellation
3. Always check nth term test first: If $\lim a_n \neq 0$, you're done (diverges)!
4. Harmonic series diverges: Memorize this fact
5. Watch starting index: $n=0$ vs $n=1$ changes the sum
6. Partial sums: When in doubt, find $S_n$ formula and take limit
7. Series vs Sequence: Series is sum, sequence is list!

$= \frac{3}{5} \sum_{n=0}^{\infty} \frac{2^n}{5^n}$

$= \frac{3}{5} \sum_{n=0}^{\infty} \left(\frac{2}{5}\right)^n$

Step 2: Identify geometric series

This is geometric with:

• First term: $a = 1$ (when summing $r^n$ from $n=0$)
• Common ratio: $r = \frac{2}{5}$

Since $|r| = \frac{2}{5} < 1$, it converges!

Step 3: Apply formula

$\sum_{n=0}^{\infty} \left(\frac{2}{5}\right)^n = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$

Step 4: Multiply by constant

$\frac{3}{5} \cdot \frac{5}{3} = 1$

Answer: The series converges to $1$.

Step 2: Expand terms

$S_n = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n+1} - \frac{1}{n+2}\right)$

Step 3: Look for cancellation

Rearrange to see pattern: $S_n = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots + \frac{1}{n+1} - \frac{1}{n+2}$

Most terms cancel!

$S_n = \frac{1}{2} - \frac{1}{n+2}$

Step 4: Take limit

$S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(\frac{1}{2} - \frac{1}{n+2}\right)$

$= \frac{1}{2} - 0 = \frac{1}{2}$

Answer: The series converges to $\frac{1}{2}$.