Comprehensive Guide to Series, Partial Sums, and Convergence Tests for Convergence, and Geometric Series Analysis of Series

Introduction to Series and Sigma Notation

• Transition from Sequences: Previously, the focus was on sequences, which are lists of terms ($a_1, a_2, a_3, \dots$). In sequences, we assess whether the list itself converges or diverges by taking the limit of the general term ($\lim_{n \to \infty} a_n$), using tools like the Squeeze Theorem or comparison tests.

• Conceptual Shift to Series: If we decide to add the terms of a sequence together rather than just list them, we create a series. This is essentially the progression from a list to a sum.

• Sigma Notation ($\Sigma$): The Capital Greek letter Sigma represents a sum. It tells us where to start adding and where to stop.

  • A typical series notation is written as $\sum_{n=1}^{\infty} a_n$, which instructs us to take the terms of the sequence and add them together forever (to infinity).

  • The starting point (index) is usually indicated at the bottom (e.g., $n=1$), and the stopping point is indicates at the top (e.g., $\infty$).

Partial Sums and the Sequence of Partial Sums

• The Partial Sum ($S_n$): Because we cannot add terms forever (the speaker notes, "I'm not going to live forever"), we look at what is called a partial sum.

• A partial sum is a finite sum that adds the first $n$ terms of a series. It is denoted as:     $S_n = \sum_{k=1}^{n} a_k$

• Existence of Partial Sums: Since $n$ is a finite number (e.g., 10, 100, or a million), the sum of these terms for sure exists. For every integers $n$, there is a specific sum $S_n$.

• Sequence of Partial Sums: For every additional term we include, we get a new sum. This creates a new sequence composed of the individual partial sums:

  • $S_1 = a_1$

  • $S_2 = a_1 + a_2$

  • $S_3 = a_1 + a_2 + a_3$

  • $S_n = a_1 + a_2 + \dots + a_n$

• This resulting sequence ($S_1, S_2, S_3, \dots, S_n$) is the "sequence of partial sums."

The Definition of Series Convergence

• Connecting Sequences to Series: The main reason sequences were studied is because the convergence of a series is defined by the convergence of its sequence of partial sums ($S_n$).

• Convergence Criterion: If the sequence of partial sums $S_n$ converges, then the limit of $S_n$ as $n$ approaches infinity exists. Let that limit be $s$:     $\lim_{n \to \infty} S_n = s$

• If this limit exists, we say the infinite series converges, and its sum is precisely $s$.

• Divergence Criterion: If the sequence of partial sums diverges (goes to infinity, negative infinity, or oscillates), then the series itself is divergent and we cannot find a finite sum.

• Calculus Jump: We use limits to equate the infinite sum with the limit of the partial sums:     $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} \sum_{k=1}^{n} a_k$

Example Study: The Series of Natural Numbers

• Problem: Determine the convergence of $\sum_{n=1}^{\infty} n$.

• Step 1: Find the partial sum ($S_n$). This is the sum of integers from 1 to $n$:     $S_n = 1 + 2 + 3 + \dots + n$

• Step 2: Find a formula for the partial sum. The speaker references the story of the mathematician "gaus" (Gaus) who found this formula in elementary school:     $S_n = \frac{n(n+1)}{2}$

• Step 3: Analyze the limit as $n \to \infty$:     $\lim_{n \to \infty} \frac{n(n+1)}{2} = \infty$

• Conclusion: Since the sequence of partial sums is divergent, the series $\sum n$ is divergent.

Telescoping Series and Partial Sum Formulas

• Definition: A telescoping series is one where internal terms cancel each other out, leaving only a few beginning and ending terms. This occurs frequently with differences of fractions.

• Example Problem: Find the sum of $\sum_{n=1}^{\infty} (\frac{1}{2n+1} - \frac{1}{2n+3})$.

  • Step 1: List the partial sum terms.

    • For $n=1$: $(\frac{1}{3} - \frac{1}{5})$

    • For $n=2$: $(\frac{1}{5} - \frac{1}{7})$

    • For $n=3$: $(\frac{1}{7} - \frac{1}{9})$

    • For $n=n-1$: $(\frac{1}{2n-1} - \frac{1}{2n+1})$

    • For $n=n$: $(\frac{1}{2n+1} - \frac{1}{2n+3})$

  • Pattern Identification: The first fraction of one term cancels with the last fraction of the previous term. Following the pattern carefully:

    • $-1/5$ cancels with $+1/5$.

    • $-1/7$ cancels with $+1/7$.

    • $-1/9$ would cancel with the next fraction.

    • $1/(2n+1)$ cancels with the preceding term's end.

  • Remaining Terms: Only the very first fraction and the very last fraction remain in the partial sum formula:         $S_n = \frac{1}{3} - \frac{1}{2n+3}$

  • Take the Limit:         $\lim_{n \to \infty} (\frac{1}{3} - \frac{1}{2n+3}) = \frac{1}{3} - 0 = \frac{1}{3}$

  • Result: The series converges to $\frac{1}{3}$.

Example Study: Decomposing Fractions for Telescoping Series

• Problem: Evaluate $\sum_{n=1}^{\infty} \frac{4}{4n^2 - 1}$.

• Analysis: This does not initially look like a telescoping series because it is not two separate fractions. We must create them.

• Step 1: Factor the denominator using the difference of squares:     $4n^2 - 1 = (2n-1)(2n+1)$

• Step 2: Perform Partial Fraction Decomposition (referred to as a "decomposition" or "decomposing the nasty fraction").     $\frac{4}{(2n-1)(2n+1)} = \frac{a}{2n-1} + \frac{b}{2n+1}$

  • Solve for $a$ and $b$: $4 = a(2n+1) + b(2n-1)$.

  • Let $n = 1/2$: $4 = a(2) \implies a = 2$.

  • Let $n = -1/2$: $4 = b(-2) \implies b = -2$.

  • Resulting Series: $\sum_{n=1}^{\infty} (\frac{2}{2n-1} - \frac{2}{2n+1})$.

• Step 3: List partial sum terms to find the pattern.

  • $n=1$: $\frac{2}{1} - \frac{2}{3}$

  • $n=2$: $\frac{2}{3} - \frac{2}{5}$

  • $n=3$: $\frac{2}{5} - \frac{2}{7}$

  • $n=n-1$: $\frac{2}{2n-3} - \frac{2}{2n-1}$

  • $n=n$: $\frac{2}{2n-1} - \frac{2}{2n+1}$

• Cancellations: The "insides" of the terms cancel out. Only the first term of the first set ($2$) and the last term of the $n$-th set ($\frac{2}{2n+1}$) remain.

• Final Formula: $S_n = 2 - \frac{2}{2n+1}$.

• Limit: $\lim_{n \to \infty} S_n = 2 - 0 = 2$.

• The series converges to $2$.

Geometric Series: Definition, Convergence, and Sum Formula

• Standard Form: A geometric series typically takes the form:     $\sum_{n=1}^{\infty} a r^{n-1} = a + ar + ar^2 + ar^3 + \dots$

  • Alternatively, it can be written starting at zero: $\sum_{n=0}^{\infty} a r^{n}$.

• Role of Constants:

  • $a$ is a constant coefficient (the first term). It doesn't affect the fact of convergence/divergence but does affect the final sum.

  • $r$ is the common ratio (the base).

• Convergence Criteria:

  • The series converges if and only if $|r| < 1$ (meaning $-1 < r < 1$). This happens only with fractions whose absolute value is less than 1. Powers of such fractions decrease toward zero as the exponent increases.

  • If $r = 1$, the series is $a + a + a + \dots$, which goes to infinity (with constant $a$).

  • The series diverges if $|r| \ge 1$ .

• The Sum Formula: If a geometric series converges, its sum is given by:     $S = \frac{a}{1-r}$

Proof of the Geometric Series Sum Formula

• Define the partial sum $S_n$ for a geometric series:     $S_n = a + ar + ar^2 + \dots + ar^{n-1}$

• Multiply the entire equation by $r$:     $r S_n = ar + ar^2 + ar^3 + \dots + ar^n$

• Subtract the two equations ($S_n - r S_n$):     $S_n (1-r) = a - ar^n$

• Solve for $S_n$:     $S_n = \frac{a(1 - r^n)}{1 - r}$

• Apply the limit as $n \to \infty$:

  • If $|r| < 1$, then $r^n \to 0$. This leaves $\frac{a(1-0)}{1-r} = \frac{a}{1-r}$.

  • If $|r| \ge 1$, the limit goes to infinity (diverges).

Analyzing Geometric Series: Convergence Criteria and Examples

• Example 1: $\sum_{n=1}^{\infty} 3 (\frac{-1}{2})^{n-1}$.

  • $a = 3$.

  • $r = -1/2$.

  • Convergence check: Since $|-1/2| < 1$, the series converges.

  • Sum: $S = \frac{3}{1 - (-1/2)} = \frac{3}{3/2} = 2$.

• Warning on Divergence: If $r = 4/3$, the sum formula will still give you a numerical value (a "silly answer"), but it is incorrect because the series is divergent. You must show convergence before calculating the sum.

• Example 2 (Pattern-based): Sum the alternating series $5/3 - 5/9 + 5/27 - 5/81 + \dots$

  • Notice it is starting at $n=1$, requiring power $n-1$.

  • Factor out the first term to ensure the internal expression starts with 1: $5/3(1 - 1/3 + 1/9 - 1/27 + \dots)$.

  • Identify common ratio: $r = -1/3$. First term $a = 5/3$.

  • Sum: $S = \frac{5/3}{1 - (-1/3)} = \frac{5/3}{4/3} = 5/4 = 1.25$.

Strategic Manipulation of Series Indices and Bases

• Problem: Evaluate $\sum_{n=1}^{\infty} \frac{e^n}{3^{n+1}}$.

• Manipulation Goal: Match the index ($n=1$) to the power ($n-1$).

  • Rewrite $e^n = e \times e^{n-1}$.

  • Rewrite $3^{n+1} = 3^2 \times 3^{n-1} = 9 \times 3^{n-1}$.

  • The series becomes: $\sum_{n=1}^{\infty} (\frac{e}{9}) (\frac{e}{3})^{n-1}$.

• Parameters: $a = e/9$ and $r = e/3$.

• Check Convergence: Since $e \approx 2.71$, the ratio $2.71/3$ is less than 1. This is a convergent series.

• Calculate Sum:     $S = \frac{e/9}{1 - e/3} = \frac{e/9}{(3-e)/3} = \frac{e}{9} \times \frac{3}{3-e} = \frac{e}{3(3-e)}$

• Algebraic Rule: When stripping off fractions to adjust powers, the ratio $r$ never changes; only the first term $a$ changes.

The Harmonic Series and Subsequence Properties

• Definition of Harmonic Series: $\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$

• Convergence status: Despite having terms that decrease toward zero, the harmonic series is divergent.

• Theorem on Subsequences:

  • If a sequence is convergent, all of its subsequences are convergent.

  • Conversely, if a subsequence is shown to be divergent, the entire sequence must be divergent.

  • The Divergence of the harmonic series is proven using subsequences of its partial sums.

The Divergence Test (n-th Term Test)

• Conditional Relationship: If a series $\sum a_n$ converges, then the limit of the individual terms must be zero ($\lim_{n \to \infty} a_n = 0$).

  • Logic: To arrive at a finite sum, eventually you must be adding nothing but zeros at infinity.

• The Divergence Test: If $\lim_{n \to \infty} a_n \neq 0$ or if the limit does not exist, then the series must diverge.

• Vital Distinction: This is not a bi-conditional statement. Having a limit equal to zero does not necessarily mean the series converges (e.g., the harmonic series terms go to 0, but it diverges).

• Example Problem: $\sum_{n=1}^{\infty} \frac{2n^2 + 1}{3n^2 - 1}$.

  • Test: Take the limit of the general term: $\lim_{n \to \infty} \frac{2n^2+1}{3n^2-1} = 2/3$.

  • Result: Since $2/3 \neq 0$, the series diverges by the Divergence Test. At infinity, you are essentially adding $2/3 + 2/3 + 2/3 \dots$ which grows without bound.

Algebraic Properties of Convergent Series

• Scalar Multiple: $\sum c a_n = c \sum a_n$. You can pull a constant out of the summation.

• Sum/Difference Rule: $\sum (a_n \pm b_n) = \sum a_n \pm \sum b_n$.

• Condition: For the sum/difference property to hold for the resulting sum, both individual series parts must be convergent. If one part is divergent, the entire combined series is divergent.

• Example Problem: $\sum_{n=1}^{\infty} \frac{2^n - 5^n}{3^n} = \sum (\frac{2}{3})^n - \sum (\frac{5}{3})^n$.

  • Right Part: Ratio $r = 5/3 > 1$. It is divergent.

  • Conclusion: Since one piece is divergent, the whole series is divergent. There is no need to calculate the sum of the first part.

Advanced Shortcut for Geometric Series Sums: The "First Term" Method

• Generalizing the sum formula: The sum of a convergent geometric series can always be calculated as:     $S = \frac{\text{First Term}}{1 - \text{Ratio}}$

• This bypasses the need to convert every series into the traditional $n-1$ or $n$ power format.

• Example Manipulation: $\sum_{n=2}^{\infty} \frac{e^n}{3^{n+1}}$.

  • Rather than shifting indices, identify the very first term by plugging the starting index ($n=2$) into the expression:

  • First Term ($a$) = $\frac{e^2}{3^{2+1}} = \frac{e^2}{27}$.

  • The Common Ratio ($r$) = $\frac{e}{3}$.

  • Sum = $\frac{e^2/27}{1 - e/3}$. This yields the same result with fewer algebraic steps.

• Crucial Requirement: For this to work, you must be absolutely sure the series is geometric (constant base to an $n$ power).

Questions & Discussion

• Question: For the telescoping example, how did the $1/(2n-3)$ part get handled?

• Response: There is a cancellation relationship where the "outsides" or certain specific fractions remain. If you have an $n$-th term, and you don't go any further, that term survives if there is no subsequent term to cancel it. We pretend we have a "dot dot dot" representing all internal terms.

• Question: Why doesn't the limit of terms to zero guarantee convergence?

• Response: The harmonic series is the primary counter-example. The terms get small enough for the sequence to converge to 0, but the total sum still grows infinitely large. The Divergence Test only identifies divergence; it never confirms convergence.

• Question: In the index shortcut, do you always just plug in whatever $n$ equals?

• Response: Yes. Whatever the starting index is ($n=0, n=1, n=2, n=4$), you plug that into the expression to find the actual first term used in the sum formula.