# Infinite Series

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# Motivation

A sequence is a list of things (usually numbers) arranged in a definite order according to some rule or pattern. Each number in the sequence is called a term, e.g., {3, 5, 7, . . . , 21}, {3, 2, 1} is 3 to 1 backwards, {a, b, c, …, z} is the sequence of the alphabetic letters, and {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} is the sequence of the first 10 prime numbers.

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 + ···.

# Infinite Series

A finite series is given by all the terms of a finite sequence, added together, e.g., {3, 5, 7, . . . , 21}, $\sum_{k=1}^{10} 2k+1 = 120$. An infinite series is the sum of an infinite sequence of numbers. It is represented in the form $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + ···$ where an represents the terms of the sequence, and n is the index that ranges from 1 to infinite.

A series is convergent (or converges) if the sequence of its partial sums tends to a limit, that is, l = $\lim_{n \to ∞} \sum_{k=1}^n a_k$ exists and is a finite number. More precisely, if there exists a number l (or S) such that for every arbitrary small positive number ε, there is a (sufficient large) N, such that ∀n ≥ N, |Sn -l| < ε where Sn = $\sum_{k=1}^n a_k = a_1 + a_2 + ··· + a_n$. If the series is convergent, the number l is called the sum of the series. On the contrary, any series that is not convergent ($\lim_{n \to ∞} \sum_{k=1}^n a_k$ does not exist) is said to be divergent or to diverge.

# Arithmetic and geometric 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. ∀ k≥ 1, ak = ak-1 + d = ak-2 + 2d = ··· = a1 + d(k-1).

The nth partial sum of an arithmetic sequence is] $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n a_1 +(k-1)d = n·a_1+d\sum_{k=1}^n (k-1) = n·a_1+d\sum_{k=2}^n (k-1) = n·a_1+d\sum_{k=1}^{n-1} k$[Sum identity, $\sum_{k=1}^n k = \frac{1}{2}n(n+1)$] = $n·a_1+d·\frac{1}{2}(n-1)n = \frac{1}{2}n(2a_1+d(n-1))$⇒[$a_1+a_n = a_1 + a_1+d(n-1)=2a_1+d(n-1)$] $S_n = \frac{n·(a_1+a_n)}{2}$.

$S_n = n·a_1+d·\frac{1}{2}(n-1)n = (a_1-\frac{d}{2})n+\frac{d}{2}n^2$. Since both quadratic and linear functions tend to infinity as n increases, the only way that an arithmetic series converges is when $\frac{d}{2}$ and $a-\frac{d}{2}$ are both equal to zero ⇒ a = d = 0, so the only arithmetic series that converges is 0 + 0 + 0 + ···

A geometric series is a specific type of infinite series where each term is obtained by multiplying the previous term by a fixed, non-zero constant. The general form of a geometric series is: $\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty ar^n = a + ar + ar^2 + ar^3 + ···$

$S_n = a + ar + ar^2 + ar^3 + ··· + ar^{n-1}$ ⇒[Multiplying both sides of the equation by r] $rS_n = ar + ar^2 + ar^3 + ar^4 + ··· + ar^{n}$ ⇒[Subtracting these equations we then obtain] $S_n -rS_n = a - ar^{n} ⇒ S_n(1-r) = a(1-r^n) ⇒$[Assume r ≠ 1, we can divide both sides by 1-r and obtain the formula for the nth partial sum of a geometric sequence] $S_n = \frac{a(1-r^n)}{1-r}$

If |r| < 1, $\lim_{n \to ∞}S_n = \lim_{n \to ∞} \frac{a(1-r^n)}{1-r} = \frac{a(1-0)}{1-r} = \frac{a}{1-r}$. Otherwise, |r| > 1, 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] = $\frac{1}{1-\frac{1}{2}} = \frac{1}{1/2} = 2.$

# 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 (Credits: Monroe_Community_College, Mathematics Libre Texts):

1. 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. 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. For any real number c, the series $\sum_{n=1}^\infty ca_n$ converges, and $\sum_{n=1}^\infty ca_n = c\sum_{n=1}^\infty a_n$.

# Examples of convergent and divergent series

• $\frac{3}{9} = 0.3 + 0.03 + 0.003 + ··· = \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + ···. \sum_{n=1}^\infty \frac{3}{10^n}$ this is a geometrical serie where a = 3/10, and r = $\frac{1}{10}< 1$, hence $\sum_{n=1}^\infty \frac{3}{10^n} = \frac{\frac{3}{10}}{1-\frac{1}{10}} = \frac{\frac{3}{10}}{\frac{9}{10}} = \frac{3}{9}.$

• 1 + 1 + 1 + 1 + ··· 1 = $\sum_{n=1}^\infty 1 = ∞$. It is a divergent series because the sum of the terms does not approach a finite value, $\lim_{n \to ∞} S_n = \lim_{n \to ∞} n = ∞$.

• $\sum_{n=1}^\infty \frac{(-3)^{n+1}}{4^{n-1}}$ this is a geometrical serie where a = $\frac{(-3)^{2}}{4^{0}}=9$, and r = $\frac{-3}{4}< 1$, hence $\sum_{n=1}^\infty \frac{(-3)^{n+1}}{4^{n-1}} = \frac{9}{1+\frac{3}{4}}=\frac{36}{7}.$

• $\sum_{n=1}^\infty \frac{1}{n(n+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+···$

The general formula for the partial sum is $s_n = \sum_{k=1}^n \frac{1}{k(k+1)}$ =[We can use partial fraction decomposition to express it as] $\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{k+1-k}{k(k+1)} = \sum_{k=1}^n \frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k}-\frac{1}{(k+1)} = (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + ··· + (\frac{1}{n-1}-\frac{1}{n}) + (\frac{1}{n}-\frac{1}{n+1})$[As the reader could easily see, most of the terms in this expression cancel in pairs.]= $1 -\frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$

$\lim_{n \to ∞}S_n = \lim_{n \to ∞} \frac{n}{n+1} = 1$. The sequence of partial sums converges and its value is $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$.

• $\sum_{n=2}^\infty (\frac{1}{n+1}-\frac{1}{n+2})$. The general formula for the partial sum is $s_n = \sum_{i=2}^n \frac{1}{n+1}-\frac{1}{n+2} = (\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4}) + (\frac{1}{4}-\frac{1}{5})+···(\frac{1}{n+1}-\frac{1}{n+2})$ =[As the reader could easily see, most of the terms in this expression cancel in pairs.] $\frac{1}{2}-\frac{1}{n+2}$

$\lim_{n \to ∞}S_n = \lim_{n \to ∞} \frac{1}{2}-\frac{1}{n+2} = \frac{1}{2}$. The sequence of partial sums converges and its value is $\sum_{n=2}^\infty (\frac{1}{n+1}-\frac{1}{n+2}) = \frac{1}{2}$

• $\sum_{n=2}^\infty \frac{1}{n^2-1}$. The general formula for the partial sum is $s_n = \sum_{i=2}^n \frac{1}{i^2-1}$ =[We can use partial fraction decomposition to express it as] $\sum_{i=2}^n \frac{1/2}{i-1} + \frac{-1/2}{i+1}= \frac{1}{2}(\frac{1}{1}-\frac{1}{3}) + \frac{1}{2}(\frac{1}{2}-\frac{1}{4}) + \frac{1}{2}(\frac{1}{3}-\frac{1}{5}) + ··· + \frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1}) = \frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+ ··· \frac{1}{n-3}-\frac{1}{n-1}+\frac{1}{n-2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}) =$[As the reader could easily see, most of the terms in this expression cancel in pairs.] $\frac{1}{2}(\frac{1}{1}+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}) = \frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$

$\lim_{n \to ∞}S_n = \lim_{n \to ∞} (\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}) = \frac{3}{4}$. The sequence of partial sums converges and its value is $\sum_{n=2}^\infty \frac{1}{n^2-1} = \frac{3}{4}$

• 1 - 1 - 1 - 1 + ···, Sn = 1 −1 +1 −1 + ···+ (−1)n. Depending on whether n is even or odd, the partial sum alternates between 0 and 1. Formally, for even n, Sn = 0, and for odd n, Sn = 1 ⇒ Sn does not converge to a fixed value as n approaches infinity, so the series 1 - 1 - 1 - 1 + ··· is divergent.

• $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3}+···$. It is a geometric series where a = 1 (the first term), r = 1/3, |r|< 1, so the series is convergent and its sum is $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}.$

• $\sum_{n=1}^\infty (\frac{3}{n(n+1)}+\frac{1}{3^{n-1}})$ is convergent because we have previously demonstrated that $\sum_{n=1}^\infty \frac{1}{3^{n-1}} = \frac{3}{2},$ and $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$ ⇒ By using the Algebraic Properties of Convergent Series, $\sum_{n=1}^\infty (\frac{3}{n(n+1)}+\frac{1}{3^{n-1}}) = 3·\sum_{n=1}^\infty (\frac{1}{n(n+1)})+ \sum_{n=1}^\infty \frac{1}{3^{n-1}} = 3·1 + \frac{3}{2} = \frac{9}{2}$.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
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, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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