# Recall

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.

Divergence Test. If $\lim_{n \to ∞}a_n ≠ 0$, then $\sum_{n=1}^\infty a_n$ diverges.

Integral Comparison. If f(x) is a positive, continuous, and decreasing function on the interval [1, ∞), then $|\sum_{n=1}^\infty f(n) -\int_{1}^{∞} f(x)dx| < f(1).$ Besides, $\sum_{n=1}^\infty f(n)$ converges, if and only if, $\int_{1}^{∞} f(x)dx$ converges.

Theorem. Direct Comparison test. Let {an} and {bn} be positive sequences where an≤bn ∀n≥N, for some N.

1. If $\sum_{n=1}^\infty b_n$ converges, then $\sum_{n=1}^\infty a_n$ converges.
2. If $\sum_{n=1}^\infty a_n$ diverges, then $\sum_{n=1}^\infty b_n$ converges.

# Ratio Test

Theorem. Ratio Test. Suppose we have the series $\sum_{n=1}^\infty a_n$, and L = $\lim_{n \to ∞} |\frac{a_{n+1}}{a_n}|$. Then, the following statements hold true,

1. If L < 1 the series $\sum_{n=1}^\infty a_n$ is absolutely convergent and hence convergent.
2. If L > 1 the series $\sum_{n=1}^\infty a_n$ is divergent.
3. If L = 1 the series may be divergent, conditionally convergent, or absolutely convergent.

# Root Test

Theorem. Root Test. Suppose we have the series $\sum_{n=1}^\infty a_n$, and L = $\lim_{n \to ∞} \sqrt[n]{|a_n|} = \lim_{n \to ∞} |a_n|^{\frac{1}{n}}$. Then,

1. If L < 1, the series is absolutely convergence ⇒ convergent.
2. If L > 1, the series is divergent.
3. If L = 1, the series may diverge, converge absolutely or converge conditionally.

Proof.

Let’s assume that L < 1 ⇒ ∃r: L < r < 1.

L = $\lim_{n \to ∞} \sqrt[n]{|a_n|} = \lim_{n \to ∞} |a_n|^{\frac{1}{n}}$ ⇒ Let ε>0 such that L + ε = r, ∃N: ∀n ≥ N, 0 ≤[This is trivial, |an| ≥ 0 ∀n ∈ ℕ]$\sqrt[n]{|a_n|}< L + ε = r$ (We know more than that, $L -ε < \sqrt[n]{|a_n|} < L + ε$)

$\sqrt[n]{|a_n|}< r ↭ |a_n|^{\frac{1}{n}} < r ⇒ |a_n| < r^n$.

Now we know that the series $\sum_{n=0}^\infty r^n$ is a convergent geometric series because 0 < r < 1 ⇒ By the comparison test, $|a_n| < r^n, \sum_{n=N}^\infty |a_n|$ converges, too.

Futhermore, since $\sum_{n=1}^\infty |a_n| = \sum_{n=1}^{N-1} |a_n| + \sum_{n=N}^\infty |a_n|$, we know that $\sum_{n=1}^\infty |a_n|$ is convergent, too since the first term on the right side of the equation is a finite sum of finite terms and therefore, finite.

Let’s now suppose that L > 1 ⇒[L = $\lim_{n \to ∞} \sqrt[n]{|a_n|} = \lim_{n \to ∞} |a_n|^{\frac{1}{n}}$] ∃N such that ∀n ≥ N: $|a_n|^{\frac{1}{n}} > 1$ (Consider ε such that L - ε = 1, $1 = L - ε < |a_n|^{\frac{1}{n}} < L + ε$).

$|a_n|^{\frac{1}{n}} > 1 ⇒ |a_n| > 1^n = 1.$ To sum up, ∀n ≥ N, |an| > 1 ⇒ $\lim_{n \to ∞} |a_n| ≠ 0 ⇒ \lim_{n \to ∞} a_n ≠ 0$ ⇒[By the Divergence Test] $\sum a_n$ is divergent.

If L = 1, the series may diverge ($\sum_{n=1}^\infty \frac{1}{n}$), converge absolutely ($\sum_{n=1}^\infty \frac{1}{n^2}$) or converge conditionally ($\sum_{n=1}^\infty \frac{(-1)^n}{n}$).

# Solved examples

• Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{n}{5^n}$.

We are going to use the Ratio Test, L = $\lim_{n \to ∞}\frac{\frac{n+1}{5^{n+1}}}{\frac{n}{5^n}} = \lim_{n \to ∞}\frac{\frac{n+1}{5}}{n} = \lim_{n \to ∞} \frac{n+1}{5n} = \frac{1}{5} < 1 ⇒ \sum_{n=1}^\infty \frac{n}{5^n}$ converges.

• Find the radius and interval of convergence of the serie $\sum_{n=1}^\infty \frac{x^n}{n}$.

Let’s use the Ratio Test, L = $\lim_{n \to ∞}\frac{\frac{|x|^{n+1}}{n+1}}{\frac{|x|^n}{n}} = \lim_{n \to ∞}\frac{|x|^{n+1}}{n+1} \frac{n}{|x|^n} = \lim_{n \to ∞} |x|\frac{n}{n+1} = |x|$ ⇒

1. If |x| < 1, the series converges.
2. If |x| > 1, the series diverges.
3. If |x| = 1 ⇒ x = ± 1. If x = 1, $\sum_{n=1}^\infty \frac{x^n}{n} = \sum_{n=1}^\infty \frac{1}{n}$ is the harmonic series and diverges. If x = -1 is the negative of the alternating harmonic series and converges. I = [-1, 1) is the interval of convergence and we may think of $\sum_{n=1}^\infty \frac{x^n}{n}$ as a function from the interval of convergence to the real numbers.
• Find the radius and interval of convergence of the serie $\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$.

Let’s use the Ratio Test, L = $\lim_{n \to ∞}|\frac{(-1)^{n+1}\frac{x^{2(n+1)}}{(2(n+1))!}}{(-1)^n\frac{x^{2n}}{(2n)!}}| = \lim_{n \to ∞}|\frac{(-1)^{n+1}\frac{x^{2}}{(2n+2)(2n+1)(2n)!}}{(-1)^n\frac{1}{(2n)!}}| = \lim_{n \to ∞}|\frac{x^{2}}{(2n+2)(2n+1)}| = \lim_{n \to ∞}\frac{x^{2}}{(2n+2)(2n+1)} = 0 < 1$, the series always converges, R = ∞, I = (-∞, +∞).

• Find the radius and interval of convergence of the serie $\sum_{n=0}^\infty \frac{(x-3)^n}{n^7+1}$.

Let’s use the Ratio Test, L = $\lim_{n \to ∞}|\frac{\frac{(x-3)^{n+1}}{(n+1)^7+1}}{\frac{(x-3)^n}{n^7+1}}| = \lim_{n \to ∞}|\frac{\frac{(x-3)}{(n+1)^7+1}}{\frac{1}{n^7+1}}| = \lim_{n \to ∞}|\frac{(x-3)(n^7+1)}{(n+1)^7+1}| = |x-3|·\lim_{n \to ∞}|\frac{(n^7+1)}{(n+1)^7+1}|$=[We can divide both the numerator and denominator by n7 and then simplify.] $|x-3|·\lim_{n \to ∞}|\frac{(1+\frac{1}{n^7})}{(\frac{n+1}{n})^7+\frac{1}{n^7}}| = |x-3|·\lim_{n \to ∞}|\frac{(1+\frac{1}{n^7})}{(1+\frac{1}{n})^7+\frac{1}{n^7}}| = |x-3|·1 = |x-3| < 1$ is required to converge. The series is centered at 3, its radius of convergence is 1, and we need to check its endpoints, namely 2 and 4.

Let x = 2, $\sum_{n=0}^\infty \frac{(x-3)^n}{n^7+1} = \sum_{n=0}^\infty \frac{(-1)^n}{n^7+1}$. We can use the alternating series test, it states that an alternating series converges if

1. bn = $\frac{1}{n^7+1}$ is a decreasing sequence, i.e., bn+1 ≤ bn ↭ $\frac{1}{(n+1)^7+1} ≤ \frac{1}{n^7+1}$
2. $\lim_{n \to ∞}b_n = 0 ↭ \lim_{n \to ∞} \frac{1}{n^7+1} = 0$ ⇒ x = 2 ∈ I.

Let x = 4, $\sum_{n=0}^\infty \frac{(x-3)^n}{n^7+1} = \sum_{n=0}^\infty \frac{(4-3)^n}{n^7+1} = \sum_{n=0}^\infty \frac{1}{n^7+1}$.

$\frac{1}{n^7+1}≤ \frac{1}{n^7}$ and $\sum_{n=1}^\infty \frac{1}{n^7}$ converges by Theorem (p-series), p = 7 > 1 ⇒[Direct Comparison Theorem] $\sum_{n=1}^\infty \frac{1}{n^7+1}$ converges ⇒ $\sum_{n=0}^\infty \frac{1}{n^7+1}$ converges ⇒ x = 4 ∈ I ⇒ I = [2, 4].

• Determine convergence or divergence of the series $\sum_{n=1}^\infty n^nx^n$.

Let’s use the Root Test.

$\lim_{n \to ∞} \sqrt[n]{|n^nx^n|} = \lim_{n \to ∞} n·\sqrt[n]{|x|^n} = \lim_{n \to ∞} n·|x| = |x|·\lim_{n \to ∞} n < 1$ only when x = 0 ⇒ R = 0 and I = {0}.

• Determine convergence or divergence of the series $\sum_{n=0}^\infty (1+\frac{1}{n})^{n^2}$.

Let’s use the Root Test.

$\lim_{n \to ∞} \sqrt[n]{(1+\frac{1}{n})^{n^2}} = \lim_{n \to ∞} (1+\frac{1}{n})^{\frac{n^2}{n}} = \lim_{n \to ∞} (1+\frac{1}{n})^n = e > 1 ⇒ \sum_{n=0}^\infty (1+\frac{1}{n})^{n^2}$ diverges.

• Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{n}{2^n}$.

Let’s use the Root Test.

$\lim_{n \to ∞} \sqrt[n]{\frac{n}{2^n}} = \lim_{n \to ∞} \frac{n^{\frac{1}{n}}}{2} = \frac{1}{2}·\lim_{n \to ∞} n^{\frac{1}{n}}$

$\lim_{n \to ∞} e^{ln(n^{\frac{1}{n}})} = e^{\lim_{n \to ∞} ln(n^{\frac{1}{n}})} = e^{\lim_{n \to ∞} \frac{1}{n}·ln(n)} = e^{\lim_{n \to ∞} \frac{ln(n)}{n}} =$[L’Hôpital, ∞/∞] $e^{\lim_{n \to ∞} \frac{\frac{1}{n}}{1}} = e^0 = 1.$

$\lim_{n \to ∞} \sqrt[n]{\frac{n}{2^n}} = \frac{1}{2}·\lim_{n \to ∞} n^{\frac{1}{n}} = \frac{1}{2}·1 = \frac{1}{2} < 1$ ⇒ $\sum_{n=1}^\infty \frac{n}{2^n}$ converges.

# 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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