Sequences and Infinite Series

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Sequences and Series

Sequences

A sequence is a list of items (usually numbers) arranged in a definite order according to a specific rule or pattern. Each number in the sequence is called a term, usually denoted as $a_n$ where $n$ is the index.

Examples of sequences:

• Finite arithmetic sequence: $\{3, 5, 7, \dots, 21\}$
• Finite countdown: $\{3, 2, 1\}$
• Alphabetical letters sequence: $\{a, b, c, \dots, z\}$
• First 10 prime numbers: $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$

Infinite Series

While a sequence is a list of numbers, a series is the sum of those numbers.

A finite series adds together the terms of a finite sequence. For example, adding the terms of $\{3, 5, 7, \dots, 21\}$ yields: $\sum_{k=1}^{10} (2k+1) = 120$

An infinite series is the sum of an infinite sequence of numbers. It is represented as: $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \dots$ where $a_n$ represents the terms of the sequence, and n is the index that ranges from 1 to infinite.

The rational number $\frac{3}{9}$ can be written as 0.33333333333… = $0.\widehat{3}$ and as an infinite series as follows $0.3 + 0.03 + 0.003 + \dots$.

 

Convergence and Divergence

A series is convergent (or converges) if the sequence of its partial sums tends to a finite limit, that is, $l = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^n a_k$ exists and is a finite number.

The $n$-th partial sum $S_n$ is the sum of the first $n$ terms: $S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n$. If the limit $L = \lim_{n \to \infty} S_n$ exists and is a finite number, the series converges, and $L$ is called the sum of the series. Formally, this means for any arbitrarily small positive number $\epsilon$, there is a sufficiently large $N$ such that for all $n \geq N$, $\vert{}S_n - L\vert{} < \epsilon$.

If this limit does not exist (e.g., it goes to infinity or oscillates forever), the series is said to be divergent.

Types of Series

Arithmetic Series

An arithmetic series is the sum of a sequence in which each term is computed from the previous one by adding or subtracting a constant $d$ (the common difference). Formally, $\forall k \ge 1, a_k = a_{k-1} + d = a_1 + (k-1)d$

The $n$-th Partial Sum Formula. To find $S_n$, we can use Gauss’s trick of adding the series to itself in reverse.

$$ \begin{aligned} S_n=\sum_{k=1}^n a_k &=[\text{An arithmetic sequence has terms } a_k = a_1 + (k-1)d] \sum_{k=1}^n \big(a_1 + (k-1)d\big) \\[2pt] &=[\text{This splits naturally into two sums: }] \sum_{k=1}^n a_1 + d\sum_{k=1}^n (k-1) \\[2pt] &[\text{Since } a_1 \text{ is constant and appears n times: }] \sum_{k=1}^n a_1 = n a_1 \\[2pt] &[\text{Shift the index in the second sum }] \sum_{k=1}^n (k-1) = \sum_{k=2}^n (k-1) = \sum_{k=1}^{n-1} k \\[2pt] &[\text{Use the known identity } \sum_{k=1}^{m} k = \frac{m(m+1)}{2}] \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2} \\[2pt] &= n a_1 + d\cdot \frac{(n-1)n}{2} \\[2pt] &= [\text{Factor out } n/2] \frac{n}{2}\big(2a_1 + d(n-1)\big) \\[2pt] &[\text{Replace (d(n-1)) with the last term. The last term of the sequence is } a_n = a_1 + (n-1)d] 2a_1 + d(n-1) = a_1 + (a_1 + d(n-1)) = a_1 + a_n \\[2pt] &= \frac{n}{2}(a_1 + a_n) \end{aligned} $$

Therefore, $\boxed{S_n = \frac{n(a_1 + a_n)}{2}}$ and $S_n = \frac{n}{2}(a_1 + a_n)$ is the classical Gauss pairing formula:number of terms × average of first and last term.

Expanding this expression gives $S_n = (a_1 - \frac{d}{2})n + \frac{d}{2}n^2$. Because this is a quadratic function of $n$, it will grow to infinity as $n \to \infty$ unless both coefficients are zero. Thus, the only arithmetic series that converges is $0 + 0 + 0 + \dots$ where $a_1 = 0$ and $d = 0$.

Geometric Series

A geometric series is formed by multiplying the previous term by a fixed, non-zero constant $r$ (the common ratio). Its general form is: $\sum_{n=0}^\infty ar^n = a + ar + ar^2 + ar^3 + \dots$

The $n$-th Partial Sum Formula. We can derive the partial sum formula elegantly:

$$ \begin{aligned} S_n &=a + ar + ar^2 + \dots + ar^{n-1} \\[2pt] &[\text{Multiply both sides by } r] rS_n = ar + ar^2 + ar^3 + \dots + ar^n \\[2pt] &[\text{Subtract the second equation from the first (all middle terms cancel out!): }] S_n - rS_n = a - ar^n \\[2pt] &S_n(1 - r) = a(1 - r^n) \\[2pt] &[\text{Assuming } r \neq 1 \text{divide by 1 - r: }] S_n = \frac{a(1 - r^n)}{1 - r} \end{aligned} $$

Convergence:

• If $\vert{}r\vert{} < 1$, then as $n \to \infty$, the term $r^n \to 0$. Therefore: $\lim_{n \to \infty} S_n = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r}$
• If $\vert{}r\vert{} \geq 1$, the limit does not exist, and the series diverges.

The geometric series converges if and only if the common ratio |r| is strictly less than 1, and the sum S of a convergent geometric series is given by the formula S = $\frac{a}{1-r}$, e.g., $\sum_{n=0}^\infty \frac{1}{2^n} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ···$= [This is a geometric series with a = 1 and r = 1/2, |r| < 1] = $\frac{1}{1-\frac{1}{2}} = \frac{1}{1/2} = 2.$

Telescoping Series

A telescoping series is a special type of series where most terms cancel out in pairs when writing the partial sum, leaving only the first and last few terms. This requires using partial fraction decomposition. See examples below.

Algebraic Properties of Convergent Series

Let $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ be convergent series. Then, the following algebraic properties hold:

1. Addition. The series $\sum_{n=1}^\infty (a_n+b_n)$ converges and $\sum_{n=1}^\infty (a_n+b_n) = \sum_{n=1}^\infty (a_n) + \sum_{n=1}^\infty (b_n)$.
2. Subtraction. The series $\sum_{n=1}^\infty (a_n-b_n)$ converges and $\sum_{n=1}^\infty (a_n+b_n) = \sum_{n=1}^\infty (a_n) - \sum_{n=1}^\infty (b_n)$.
3. Scalar Multiplication. For any real number c, the series $\sum_{n=1}^\infty ca_n$ converges, and $\sum_{n=1}^\infty ca_n = c\sum_{n=1}^\infty a_n$.

Solved Examples

Geometric Series. Repeating Decimals

Convert $0.\widehat{3}$ into a fraction.

$0.333\dots = 0.3 + 0.03 + 0.003 + \dots = \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \dots$

This is a geometric series with $a = \frac{3}{10}$ and $r = \frac{1}{10}$. Since $\vert{}r\vert{} < 1$, it converges: $S = \frac{a}{1 - r} = \frac{3/10}{1 - 1/10} = \frac{3/10}{9/10} = \mathbf{\frac{3}{9}}$

Geometric Series. Standard Geometric:

Evaluate $\sum_{n=1}^\infty \frac{(-3)^{n+1}}{4^{n-1}}$

1. To find $a$, plug in $n=1$: $a = \frac{(-3)^2}{4^0} = 9$.
2. To find $r$, look at the ratio between terms: $r = -\frac{3}{4}$.
3. Since $\vert{}r\vert{} < 1$, the series converges: $S = \sum_{n=1}^\infty \frac{(-3)^{n+1}}{4^{n-1}} = \frac{9}{1 - (-3/4)} = \frac{9}{1 + 3/4} = \frac{9}{7/4} = \mathbf{\frac{36}{7}}$

Telescoping Series, Simple Telescoping, $\sum_{n=1}^\infty \frac{1}{n(n+1)}$

1. First, use partial fraction decomposition: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
2. Let’s expand the partial sum $S_n$: $S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$
3. Notice how the inner terms cancel out ($-\frac{1}{2}$ cancels with $+\frac{1}{2}$, $-\frac{1}{3}$ cancels with $\frac{1}{3}$, etc.). This leaves: $S_n = \frac{1}{1} - \frac{1}{n+1} = 1 - \frac{1}{n+1}$
4. Take the limit as $n \to \infty$: $\lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = \mathbf{1}$

Telescoping Series, Offset Telescoping, $\sum_{n=2}^\infty \frac{1}{n^2-1}$

1. Factor the denominator and use partial fractions: $\frac{1}{(n-1)(n+1)} = \frac{1/2}{n-1} - \frac{1/2}{n+1}$.
2. Let’s pull out the $\frac{1}{2}$ and write out $S_n$: $S_n = \frac{1}{2} \left[ \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \dots + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) \right]$
3. Because the denominators are spaced by $2$, it takes two steps for terms to start canceling (the $-\frac{1}{3}$ cancels with the $+\frac{1}{3}$ two terms later). At the end, the first two positive terms ($\frac{1}{1}$ and $\frac{1}{2}$) and the last two negative terms ($-\frac{1}{n}$ and $-\frac{1}{n+1}$) survive: $S_n = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1} \right) = \frac{3}{4} - \frac{1}{2n} - \frac{1}{2(n+1)}$
4. Take the limit: $\lim_{n \to \infty} S_n = \mathbf{\frac{3}{4}}$

Telescoping Series, $\sum_{n=2}^\infty (\frac{1}{n+1}-\frac{1}{n+2})$.

1. A telescoping series has the form $\sum_{n=1}^{N} (a_n - a_{n+1})$ whose partial sum collapses to $a_1 - a_{N+1}$. This is the defining property of telescoping sums: all middle terms cancel, leaving only the first and last term.
   This series fits this pattern with $a_n = \frac{1}{n+1}$.
2. Compute the partial sum.
   Write out the first few terms: $s_n = \sum_{i=2}^n \frac{1}{n+1}-\frac{1}{n+2} = (\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4}) + (\frac{1}{4}-\frac{1}{5})+···(\frac{1}{n+1}-\frac{1}{n+2})$ =[As the reader could easily see, everything cancels except the first positive term and the last negative term] $\frac{1}{2}-\frac{1}{n+2}$.
3. Take the limit. $\lim_{n \to \infty}S_n = \lim_{n \to ∞} \frac{1}{2}-\frac{1}{n+2} = \frac{1}{2}$. The sequence of partial sums converges and its value is $\sum_{n=2}^\infty (\frac{1}{n+1}-\frac{1}{n+2}) = \frac{1}{2}$

Linear Combinations, $\sum_{n=1}^\infty \left( \frac{3}{n(n+1)} + \frac{1}{3^{n-1}} \right)$

1. We can split this into two series and solve them using properties of convergent series: $\sum_{n=1}^\infty \left( \frac{3}{n(n+1)} + \frac{1}{3^{n-1}} \right) = 3 \cdot \sum_{n=1}^\infty \frac{1}{n(n+1)} + \sum_{n=1}^\infty \frac{1}{3^{n-1}}$
2. From Telescoping Series, Simple Telescoping, the first sum evaluates to $1$.
3. The second sum is a geometric series with $a=1, r=1/3$. Its sum is $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}.$
4. $\sum_{n=1}^\infty \left( \frac{3}{n(n+1)} + \frac{1}{3^{n-1}} \right) = 3(1) + \frac{3}{2} = \mathbf{\frac{9}{2}}$

Divergent Series, Grandi’s Series

1. Consider the series $\sum_{n=0}^\infty (-1)^n = 1 - 1 + 1 - 1 + 1 \dots$
2. The sequence of partial sums alternates: $S_1 = 1$; $S_2 = 1 - 1 = 0$; $S_3 = 1 - 1 + 1 = 1$
3. Because $S_n$ oscillates continuously between $0$ and $1$, the limit does not exist. The series diverges.

Summing Constants, $1 + 1 + 1 + 1 + \dots$

This is $\sum_{n=1}^\infty 1$. The $n$-th partial sum is simply $S_n = n$.

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

Because it does not approach a finite value, the series diverges.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.