Series – The Basics for MATH 141
Exam Relevance for MATH 141
Series notation in MATH 141 is setup for convergence tests. Rarely standalone questions.
What is a Series?
A series is what you get when you add up the terms of a sequence.
If you have a sequence $a_1, a_2, a_3, \ldots$, the corresponding series is:
$$a_1 + a_2 + a_3 + \cdots$$
Sigma Notation
We use the Greek letter sigma ($\Sigma$) to write series compactly:
$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$$
Reading the notation:
$$\sum_{n=1}^{\infty} a_n$$
- $\Sigma$ = "sum"
- $n = 1$ = starting index (start at $n = 1$)
- $\infty$ = ending index (continue forever)
- $a_n$ = the terms being added
Example:
$$\sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$$
Finite vs Infinite Series
Finite series: Has a definite number of terms
$$\sum_{n=1}^{5} n^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$$
Infinite series: Continues forever
$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$$
The big question for infinite series: Does the sum approach a finite number, or does it blow up to infinity (or fail to settle)?
Partial Sums
Since we can't literally add infinitely many numbers, we define series through partial sums.
The $n$th partial sum $S_n$ is the sum of the first $n$ terms:
$$S_1 = a_1$$ $$S_2 = a_1 + a_2$$ $$S_3 = a_1 + a_2 + a_3$$ $$S_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^{n} a_k$$
The infinite series equals the limit of the partial sums:
$$\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n$$
Problem: For the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$, write out $S_1$, $S_2$, $S_3$, and $S_4$.
First, identify the terms: $a_n = \dfrac{1}{n(n+1)}$
$$a_1 = \frac{1}{1 \cdot 2} = \frac{1}{2}, \quad a_2 = \frac{1}{2 \cdot 3} = \frac{1}{6}, \quad a_3 = \frac{1}{3 \cdot 4} = \frac{1}{12}, \quad a_4 = \frac{1}{4 \cdot 5} = \frac{1}{20}$$
Now build the partial sums:
$$S_1 = \frac{1}{2}$$
$$S_2 = \frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$$
$$S_3 = \frac{2}{3} + \frac{1}{12} = \frac{8}{12} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$$
$$S_4 = \frac{3}{4} + \frac{1}{20} = \frac{15}{20} + \frac{1}{20} = \frac{16}{20} = \frac{4}{5}$$
$$\boxed{S_1 = \frac{1}{2}, \quad S_2 = \frac{2}{3}, \quad S_3 = \frac{3}{4}, \quad S_4 = \frac{4}{5}}$$
Notice the pattern: $S_n = \dfrac{n}{n+1}$. As $n \to \infty$, $S_n \to 1$, so the series converges to 1.
Expanding Series from Sigma Notation
Problem: Write out the first four terms of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$.
Note this starts at $n = 0$!
$$n = 0: \quad \frac{(-1)^0}{0!} = \frac{1}{1} = 1$$
$$n = 1: \quad \frac{(-1)^1}{1!} = \frac{-1}{1} = -1$$
$$n = 2: \quad \frac{(-1)^2}{2!} = \frac{1}{2}$$
$$n = 3: \quad \frac{(-1)^3}{3!} = \frac{-1}{6}$$
$$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \cdots}$$
(This series actually converges to $e^{-1} = \frac{1}{e}$!)
Problem: Write out the first four terms of $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$.
This starts at $n = 2$ (since $\ln 1 = 0$ would cause division by zero).
$$n = 2: \quad \frac{1}{2 \ln 2}$$
$$n = 3: \quad \frac{1}{3 \ln 3}$$
$$n = 4: \quad \frac{1}{4 \ln 4}$$
$$n = 5: \quad \frac{1}{5 \ln 5}$$
$$\boxed{\sum_{n=2}^{\infty} \frac{1}{n \ln n} = \frac{1}{2 \ln 2} + \frac{1}{3 \ln 3} + \frac{1}{4 \ln 4} + \frac{1}{5 \ln 5} + \cdots}$$
Writing Series in Sigma Notation
Problem: Write in sigma notation: $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \cdots$
Identify the pattern:
- $\frac{1}{3} = \frac{1}{3^1}$
- $\frac{1}{9} = \frac{1}{3^2}$
- $\frac{1}{27} = \frac{1}{3^3}$
- $\frac{1}{81} = \frac{1}{3^4}$
The general term is $\dfrac{1}{3^n}$.
$$\boxed{\sum_{n=1}^{\infty} \frac{1}{3^n}}$$
Or equivalently: $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n$
Problem: Write in sigma notation: $1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \cdots$
The denominators are $1, 2, 3, 4, 5, \ldots$ → just $n$
The signs alternate: $+, -, +, -, +, \ldots$
Since it starts positive, use $(-1)^{n+1}$:
$$\boxed{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}}$$
(This is the famous alternating harmonic series, which converges to $\ln 2$.)
Problem: Write in sigma notation: $\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} + \cdots$
Recognize the denominators: $2 = 2!$, $6 = 3!$, $24 = 4!$, $120 = 5!$
The general term is $\dfrac{1}{n!}$ starting at $n = 2$.
$$\boxed{\sum_{n=2}^{\infty} \frac{1}{n!}}$$
Computing Finite Sums
Problem: Evaluate $\displaystyle\sum_{k=1}^{4} (2k - 1)$.
Write out and add each term:
$$k = 1: \quad 2(1) - 1 = 1$$ $$k = 2: \quad 2(2) - 1 = 3$$ $$k = 3: \quad 2(3) - 1 = 5$$ $$k = 4: \quad 2(4) - 1 = 7$$
$$\sum_{k=1}^{4} (2k - 1) = 1 + 3 + 5 + 7 = \boxed{16}$$
(Note: This is the sum of the first 4 odd numbers, which equals $4^2 = 16$!)
Problem: Evaluate $\displaystyle\sum_{i=0}^{3} 2^i$.
$$i = 0: \quad 2^0 = 1$$ $$i = 1: \quad 2^1 = 2$$ $$i = 2: \quad 2^2 = 4$$ $$i = 3: \quad 2^3 = 8$$
$$\sum_{i=0}^{3} 2^i = 1 + 2 + 4 + 8 = \boxed{15}$$
Index Shifting
The same series can be written with different starting indices. This is called index shifting or re-indexing.
Problem: Show that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ and $\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n+1)^2}$ are the same series.
Expand the first series: $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$
Expand the second series: $$\sum_{n=0}^{\infty} \frac{1}{(n+1)^2} = \frac{1}{(0+1)^2} + \frac{1}{(1+1)^2} + \frac{1}{(2+1)^2} + \cdots = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$
$$\boxed{\text{Same series, different notation}}$$
The substitution: Let $m = n + 1$. When $n = 0$, $m = 1$. Replace $n$ with $m - 1$.
Properties of Series
If $\sum a_n$ and $\sum b_n$ both converge, then:
1. Constant Multiple: $$\sum_{n=1}^{\infty} c \cdot a_n = c \cdot \sum_{n=1}^{\infty} a_n$$
2. Sum/Difference: $$\sum_{n=1}^{\infty} (a_n \pm b_n) = \sum_{n=1}^{\infty} a_n \pm \sum_{n=1}^{\infty} b_n$$
⚠️ Warning: These only work when the series converge!
Problem: If $\displaystyle\sum_{n=1}^{\infty} a_n = 3$ and $\displaystyle\sum_{n=1}^{\infty} b_n = 7$, find $\displaystyle\sum_{n=1}^{\infty} (2a_n - b_n)$.
$$\sum_{n=1}^{\infty} (2a_n - b_n) = 2\sum_{n=1}^{\infty} a_n - \sum_{n=1}^{\infty} b_n = 2(3) - 7 = \boxed{-1}$$
Common Finite Sum Formulas
These are useful for computing partial sums:
| Sum | Formula |
|---|---|
| $\displaystyle\sum_{k=1}^{n} k = 1 + 2 + \cdots + n$ | $\dfrac{n(n+1)}{2}$ |
| $\displaystyle\sum_{k=1}^{n} k^2$ | $\dfrac{n(n+1)(2n+1)}{6}$ |
| $\displaystyle\sum_{k=1}^{n} k^3$ | $\left[\dfrac{n(n+1)}{2}\right]^2$ |
Problem: Evaluate $\displaystyle\sum_{k=1}^{100} k$.
Using the formula:
$$\sum_{k=1}^{100} k = \frac{100(101)}{2} = \frac{10100}{2} = \boxed{5050}$$
This is the famous result Gauss supposedly discovered as a child!
Common Mistakes and Misunderstandings
❌ Mistake: Confusing the index variable
Wrong: Thinking $\displaystyle\sum_{n=1}^{5} n^2$ and $\displaystyle\sum_{k=1}^{5} k^2$ are different.
Why it's wrong: The index is a "dummy variable" — its name doesn't matter. Both equal $1 + 4 + 9 + 16 + 25 = 55$.
Correct: The letter used for the index ($n$, $k$, $i$, $j$) doesn't change the sum.
❌ Mistake: Forgetting to check the starting index
Wrong: For $\displaystyle\sum_{n=0}^{\infty} x^n$, saying the first term is $x$.
Why it's wrong: When $n = 0$, the first term is $x^0 = 1$, not $x$.
Correct: Always plug in the starting index to find the first term. Here: $x^0 = 1$.
❌ Mistake: Thinking $\sum a_n$ equals $a_n$
Wrong: "$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{n^2}$"
Why it's wrong: The sum $\sum$ adds up ALL the terms. $a_n = \frac{1}{n^2}$ is just ONE term.
Correct: $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots = \frac{\pi^2}{6}$
❌ Mistake: Adding infinite series like finite sums
Wrong: "I'll just add up all the terms $1 + 2 + 3 + 4 + \cdots$ and get the answer."
Why it's wrong: You can't literally add infinitely many numbers. Infinite series are defined as limits of partial sums.
Correct: $\displaystyle\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n$ where $S_n$ is the $n$th partial sum.
Infinite Series as Limit of Partial Sums
An infinite series is defined as the limit of its partial sums. If this limit exists and is finite, the series converges. Otherwise, it diverges.
Variables:
- $a_n$:
- the nth term of the series
- $S_N$:
- the Nth partial sum (sum of first N terms)
Sum of First n Positive Integers
The sum of the first n positive integers. Useful for computing partial sums and evaluating series.
Variables:
- $n$:
- the number of terms
- $k$:
- the index variable
Sum of First n Squares
The sum of the squares of the first n positive integers.
Variables:
- $n$:
- the number of terms
- $k$:
- the index variable
Linearity of Convergent Series
If both series converge, you can factor out constants and split sums. This only works for convergent series!
Variables:
- $c, d$:
- constants
- $a_n, b_n$:
- terms of convergent series
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