Convergence/Divergence of Series for MATH 101 A
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When Does an Infinite Sum Make Sense?
Adding infinitely many numbers sounds impossible — and sometimes it is! The series $1 + 2 + 3 + 4 + \cdots$ grows forever. But $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ adds up to exactly 1. The difference? Convergence.
A series converges if its partial sums approach a finite number. It diverges if they don't.
The Definition
Given a series $\displaystyle\sum_{n=1}^{\infty} a_n$, let $S_N$ be the $N$th partial sum:
$$S_N = a_1 + a_2 + \cdots + a_N$$
- If $\displaystyle\lim_{N \to \infty} S_N = L$ (a finite number), the series converges to $L$
- If the limit doesn't exist or is infinite, the series diverges
The Divergence Test (nth Term Test)
This is the simplest and most important first check for any series.
Theorem: If $\displaystyle\sum_{n=1}^{\infty} a_n$ converges, then $\displaystyle\lim_{n \to \infty} a_n = 0$.
Contrapositive (the useful form):
$$\text{If } \lim_{n \to \infty} a_n \neq 0, \text{ then } \sum a_n \text{ diverges.}$$
⚠️ Critical Warning
The Divergence Test can only prove divergence, never convergence!
If $\lim_{n \to \infty} a_n = 0$, you learn nothing — the series might converge or diverge.
Example: $\displaystyle\sum \frac{1}{n}$ has $\lim_{n \to \infty} \frac{1}{n} = 0$, but the series diverges (it's the harmonic series).
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2n+1}$ converges or diverges.
Apply the Divergence Test. Find $\displaystyle\lim_{n \to \infty} a_n$:
$$\lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2}$$
Since $\displaystyle\lim_{n \to \infty} a_n = \frac{1}{2} \neq 0$:
$$\boxed{\text{The series diverges by the Divergence Test}}$$
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} (-1)^n$ converges or diverges.
$$\lim_{n \to \infty} (-1)^n \text{ does not exist}$$
The terms alternate between $-1$ and $1$ forever.
Since the limit doesn't exist (and certainly isn't 0):
$$\boxed{\text{The series diverges by the Divergence Test}}$$
Problem: Use the Divergence Test on $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$.
$$\lim_{n \to \infty} \frac{1}{n^2} = 0$$
Since the limit equals 0, the Divergence Test is inconclusive.
$$\boxed{\text{The Divergence Test tells us nothing — need another test}}$$
(This series actually converges to $\frac{\pi^2}{6}$, but we need other methods to show it.)
Finding the Sum Using Partial Sums
When you can find a formula for $S_N$, you can determine convergence by taking the limit.
Problem: Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$, or show it diverges.
Step 1: Use partial fractions
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
Step 2: Write out the partial sum
$$S_N = \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$
$$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)$$
Step 3: Notice the telescoping!
Most terms cancel:
$$S_N = 1 - \frac{1}{N+1}$$
Step 4: Take the limit
$$\lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right) = 1 - 0 = 1$$
$$\boxed{\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1}$$
Problem: If $S_N = \dfrac{3N}{N+2}$, find $\displaystyle\sum_{n=1}^{\infty} a_n$.
The sum equals the limit of the partial sums:
$$\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{3N}{N+2}$$
Divide top and bottom by $N$:
$$= \lim_{N \to \infty} \frac{3}{1 + \frac{2}{N}} = \frac{3}{1 + 0} = 3$$
$$\boxed{\sum_{n=1}^{\infty} a_n = 3}$$
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} 1$ converges or diverges.
The partial sums are:
$$S_N = 1 + 1 + 1 + \cdots + 1 = N$$
$$\lim_{N \to \infty} S_N = \lim_{N \to \infty} N = \infty$$
$$\boxed{\text{The series diverges (to } \infty\text{)}}$$
Finding $a_n$ from $S_n$
Sometimes you're given the partial sum formula and need to find the terms.
Key relationship:
$$a_n = S_n - S_{n-1} \quad \text{for } n \geq 2$$
$$a_1 = S_1$$
Problem: If $S_n = 3n^2 + 2n$, find $a_n$ and $a_5$.
For $n \geq 2$:
$$a_n = S_n - S_{n-1}$$
$$= (3n^2 + 2n) - (3(n-1)^2 + 2(n-1))$$
$$= 3n^2 + 2n - 3(n^2 - 2n + 1) - 2n + 2$$
$$= 3n^2 + 2n - 3n^2 + 6n - 3 - 2n + 2$$
$$= 6n - 1$$
Check $a_1$:
$$S_1 = 3(1)^2 + 2(1) = 5$$
Using our formula: $a_1 = 6(1) - 1 = 5$ ✓
So $a_n = 6n - 1$ works for all $n \geq 1$.
Find $a_5$:
$$a_5 = 6(5) - 1 = 29$$
$$\boxed{a_n = 6n - 1, \quad a_5 = 29}$$
Properties of Convergent Series
If $\displaystyle\sum a_n = A$ and $\displaystyle\sum b_n = B$ (both converge), then:
| Property | Result |
|---|---|
| $\displaystyle\sum c \cdot a_n$ | $= c \cdot A$ |
| $\displaystyle\sum (a_n + b_n)$ | $= A + B$ |
| $\displaystyle\sum (a_n - b_n)$ | $= A - B$ |
Warning: These only apply when both series converge!
Problem: If $\displaystyle\sum_{n=1}^{\infty} a_n = 4$ and $\displaystyle\sum_{n=1}^{\infty} b_n = -2$, find $\displaystyle\sum_{n=1}^{\infty} (3a_n + 5b_n)$.
$$\sum_{n=1}^{\infty} (3a_n + 5b_n) = 3\sum_{n=1}^{\infty} a_n + 5\sum_{n=1}^{\infty} b_n$$
$$= 3(4) + 5(-2) = 12 - 10 = \boxed{2}$$
Adding or Removing Finite Terms
Adding or removing a finite number of terms doesn't change whether a series converges or diverges (though it changes the sum).
$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + \cdots + a_k + \sum_{n=k+1}^{\infty} a_n$$
If one side converges, so does the other.
Problem: If $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$, find $\displaystyle\sum_{n=3}^{\infty} \frac{1}{n(n+1)}$.
The series starting at $n = 3$ is missing the first two terms:
$$\sum_{n=3}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n(n+1)} - \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3}$$
$$= 1 - \frac{1}{2} - \frac{1}{6} = 1 - \frac{3}{6} - \frac{1}{6} = 1 - \frac{4}{6} = \frac{2}{6} = \boxed{\frac{1}{3}}$$
Quick Summary: Convergence vs Divergence
| If you observe... | Conclusion |
|---|---|
| $\lim_{n \to \infty} a_n \neq 0$ | Diverges (Divergence Test) |
| $\lim_{n \to \infty} a_n = 0$ | Inconclusive — need other tests |
| $\lim_{N \to \infty} S_N = L$ (finite) | Converges to $L$ |
| $\lim_{N \to \infty} S_N = \pm\infty$ | Diverges |
| $\lim_{N \to \infty} S_N$ doesn't exist | Diverges |
Common Mistakes and Misunderstandings
❌ Mistake: Using the Divergence Test to prove convergence
Wrong: "$\lim_{n \to \infty} \frac{1}{n} = 0$, so by the Divergence Test, $\sum \frac{1}{n}$ converges."
Why it's wrong: The Divergence Test can ONLY prove divergence. When the limit is 0, the test tells you absolutely nothing.
Correct: When $\lim a_n = 0$, say "The Divergence Test is inconclusive" and use another test.
❌ Mistake: Confusing $a_n \to 0$ with $\sum a_n$ converging
Wrong: "The terms get smaller, so the series must converge."
Why it's wrong: The harmonic series $\sum \frac{1}{n}$ has terms going to 0, but it diverges! Terms approaching 0 is necessary but not sufficient for convergence.
Correct: $a_n \to 0$ is required for convergence, but you need additional tests to confirm it.
❌ Mistake: Forgetting $a_1 = S_1$ when finding terms
Wrong: Using $a_n = S_n - S_{n-1}$ for $n = 1$.
Why it's wrong: $S_0$ is not defined (there's no "sum of zero terms" in most contexts). The formula only works for $n \geq 2$.
Correct: Always find $a_1$ directly as $a_1 = S_1$, then use $a_n = S_n - S_{n-1}$ for $n \geq 2$.
❌ Mistake: Thinking convergent series can be rearranged freely
Wrong: "Since $\sum a_n = L$, I can add the terms in any order and still get $L$."
Why it's wrong: This is only true for absolutely convergent series. For conditionally convergent series (like the alternating harmonic series), rearranging terms can give ANY sum you want — or even diverge!
Correct: Be careful with rearrangements. For now, keep terms in their original order.
Series Convergence Definition
A series converges to L if the limit of its partial sums equals L. If this limit doesn't exist or is infinite, the series diverges.
Variables:
- $S_N$:
- the Nth partial sum: S_N = a_1 + a_2 + ... + a_N
- $L$:
- the sum of the series (a finite number)
The Divergence Test (nth Term Test)
If the terms don't approach 0, the series diverges. WARNING: If the limit IS 0, this test is inconclusive — the series might converge or diverge!
Variables:
- $a_n$:
- the nth term of the series
Finding Terms from Partial Sums
To find the nth term when given a partial sum formula, subtract consecutive partial sums. Remember to find a_1 separately since S_0 is undefined.
Variables:
- $a_n$:
- the nth term
- $S_n$:
- partial sum of first n terms
- $S_{n-1}$:
- partial sum of first (n-1) terms
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