Opportunity often comes disguised in the form of misfortune, or temporary defeat, Napoleon Hill

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 a_{n} 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, |S_{n} -l| < ε where S_{n} = $\sum_{k=1}^n a_k = a_1 + a_2 + ··· + a_n$. 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 {a_{n}} and {b_{n}} be positive sequences where a_{n}≤b_{n} ∀n≥N, for some N.

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

A geometric series is the sum of an infinite number of terms that have a constant ratio between successive terms. It is written as a + ar + ar^{2} + ar^{3} + ··· where a is the coefficient of each term and r is the common ratio between adjacent terms.

Recall that the sum of a geometric series $\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$ if |r|< 1, and that the series diverges when |r| ≥ 1. We typically think of r as a constant, but we could also think of it as a variable, in which case the series $\sum_{n=0}^\infty ax^n$ is a function and f(x) = $\sum_{n=0}^\infty ax^n = \frac{a}{1-x}$ as long as |x| < 1.

Definition. A series of the form $\sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + ···$ where the coefficients a_{n} are real numbers is a power series centered at x = 0. A series of the form $\sum_{n=0}^\infty a_n(x-a)^n = a_0 + a_1(x-a) + a_2(x-a)^2 + ···$ where the coefficients a_{n} are real numbers is a power series centered at x = a.

Since the terms in a power series involve a variable x, the series may converge for certain values of x and diverge for other values of x. Besides, a power series centered at x = a always converges at its center (the value of its terms is a_{0}). Most power series, converge for more than one value of x.

Theorem. Convergence of a Power Series. Consider the power serie $\sum_{n=0}^\infty a_n(x-a)^n$. It satisfies exactly one of the following properties.

- The series converges at x = a and diverges ∀x ≠ a.
- The series converges for all real numbers.
- There exist a real number R (radius of convergence)> 0 such that the series converges ∀x such that |x-a|< R, and diverges ∀x such that |x-a| ≥ R.

Proof.

For the sake of simplicity let’s assume that the power series is centered at a = 0 (Otherwise, consider y = x -a, $\sum_{n=0}^\infty a_ny^n$).

Suppose ∃d ≠ 0 such that $\sum_{n=0}^\infty a_nd^n$ converges, we need to prove that $\sum_{n=0}^\infty a_nx^n$ converges ∀x, |x| < |d| (😄).

$\sum_{n=0}^\infty a_nd^n$ converges ⇒[Necessary condition for the convergence of a series.] $\lim_{n \to ∞} a_nd^n = 0$ ⇒ ∃N: $|a_nd^n|≤ 1$ ∀n ≥ N (🚀)

$|a_nx^n| = |a_nd^n|(|\frac{x}{d}|)^n$ ⇒(🚀) $|a_nx^n| ≤ (|\frac{x}{d}|)^n$

Now consider that the series $\sum_{n=N}^\infty (|\frac{x}{d}|)^n$ is a geometric series that converges if $|\frac{x}{d}|< 1$ (😄) ⇒ $\sum_{n=N}^\infty (|\frac{x}{d}|)^n$ converges ⇒[By the Comparison Test] $\sum_{n=N}^\infty a_nx^n$ converges ∀x, |x| < |d|. Since we can add a finite number of terms to a convergent series and still does not change its convergence, $\sum_{n=0}^\infty a_nx^n$ converges ∀x, |x| < |d|.

Let S be the set of real number for which the series $\sum_{n=0}^\infty a_nx^n$ converges. If S = {0} ⇒ we are in the first case and we are done. If S = ℝ ⇒ we are in the second case and we are also done.

Otherwise, suppose S ≠ {0} and S ≠ ℝ, there exist a real number “s” ≠ 0 such that the series does not converge ⇒ the series cannot converge ∀x, |x| > |s| (otherwise, it will contradict the result that we have just proven) ⇒ S must be a bounded set ⇒[By the Least Upper Bound Property] there is a smallest upper bound, let’s name it R, R > 0, and the series converges ∀x, |x| < |R| (case 3).∎

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

Let’s use the Ratio Test, L = $\lim_{n \to ∞} |\frac{\frac{(-1)^{n+1}x^{n+1}}{n+1+1}}{\frac{(-1)^nx^n}{n+1}}| = \lim_{n \to ∞} |\frac{x^{n+1}}{n+1+1}\frac{n+1}{x^n}|=\lim_{n \to ∞} |x||\frac{n+1}{n+2}|=|x|$. If |x| < 1, the power series converges, so R = 1. If |x| > 1, the power series diverges. |x| = 1 ⇒x = ± 1, we need to study it.

Let x = -1. $\sum_{n=0}^\infty \frac{(-1)^nx^n}{n+1} = \sum_{n=0}^\infty \frac{(-1)^n(-1)^n}{n+1} = \sum_{n=0}^\infty \frac{1}{n+1} = \sum_{n=1}^\infty \frac{1}{n}$, the harmonic series and we know that it diverges, x = -1 ∉ I.

Let x = 1. $\sum_{n=0}^\infty \frac{(-1)^n1^n}{n+1} = \sum_{n=0}^\infty \frac{(-1)^n }{n+1}$ ⇒[Alternating Series Test. An alternating series converges if |a_{n}| decreases monotonically and $\lim_{n \to ∞} a_n = 0$.] $\sum_{n=0}^\infty \frac{(-1)^n }{n+1}$ converges ⇒ x = 1 ∈ I, hence I = (-1, 1].

- In general, we can apply the same ideas to an arbitrary power series, $\sum_{n=0}^\infty a_nx^n$

$\lim_{n \to ∞} |\frac{a_{n+1}x^{n+1}}{a_nx^n}| = \lim_{n \to ∞} |x|\frac{|a_{n+1}|}{|a_n|} = L|x|$ when we are obviously assuming that such limit exists, i.e., L = $\lim_{n \to ∞} \frac{|a_{n+1}|}{|a_n|}$

- If L ∈ (0, ∞), the series converges if L|x| < 1 ↭ |x| < 1/L, and diverges if |x| > 1/L. Therefore, 1/L is the radius of convergence, and x = ±1/L require further investigation.
- If L = 0 ⇒ $\lim_{n \to ∞} |\frac{a_{n+1}x^{n+1}}{a_nx^n}| = L|x| = 0$ ∀x ∈ ℝ ⇒ The series converges for all x and the function is defined for all real numbers.
- If L = ∞ ⇒ $\lim_{n \to ∞} |\frac{a_{n+1}x^{n+1}}{a_nx^n}| = L|x| = ∞.$ The series converges only when x = 0.

- Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{(-1)^n·3^{n+2}}{(n-1)^n}$

$\lim_{n \to ∞} |a_n|^{\frac{1}{n}} = \lim_{n \to ∞} |\frac{(-1)^n·3^{n+2}}{(n-1)^n}|^{\frac{1}{n}} = \lim_{n \to ∞} (\frac{3^2·3^n}{(n-1)^n})^{\frac{1}{n}} = \lim_{n \to ∞} 9^{\frac{1}{n}}·\lim_{n \to ∞} \frac{3}{n-1} = 9^0·0=1·0 = 0 < 1$ ⇒[By the Root Test] $\sum_{n=1}^\infty \frac{(-1)^n·3^{n+2}}{(n-1)^n}$ converges.

- Find a power representation for f(x) = $\frac{3}{1-x^4}$ and determine the interval of confidence.

Recall $\sum_{n=0}^\infty x^n = x^0 + x^1 + x^2 + x^3 + ··· = \frac{1}{1-x}$ where |x| < 1.

f(x) = $\frac{3}{1-x^4} =3·\frac{1}{1-(x^4)}$ is already in that form, therefore f(x) = 3$\sum_{n=0}^\infty x^{4n} = \sum_{n=0}^\infty 3x^{4n} = \frac{3}{1-x^4}$ where |r| = |x^{4}| < 1 ↭ x^{4} < 1 ⇒ -1 < x < 1, hence I = (-1, 1).

- Find a power representation for f(x) = $\frac{x}{2x^2+1}$ and determine the interval of confidence.

f(x) = $\frac{x}{2x^2+1} = x\frac{1}{1+2x^2} = x\frac{1}{1-(-2x^2)} = x·\sum_{n=0}^\infty (-2x^2)^n = \sum_{n=0}^\infty (-1)^n2^{2n}x^{2n+1}$

where |r| = |-2x^{2}| < 1 ↭ |-2||x^{2}| < 1 ↭ 2·x^{2} < 1 ↭ x^{2} < ^{1}⁄_{2}, hence I = ($\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$).

- Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{n^2·2^{n-1}}{(-5)^n}$

Let’s use Radio Test,

$\lim_{n \to ∞} |\frac{a_{n+1}}{a_n}| = \lim_{n \to ∞} |\frac{\frac{(n+1)^2·2^{n}}{(-5)^{n+1}}}{\frac{n^2·2^{n-1}}{(-5)^n}}| = \lim_{n \to ∞} |\frac{\frac{(n+1)^2}{(-5)}}{n^2·2^{-1}}| = \lim_{n \to ∞} |\frac{2(n+1)^2}{(-5)n^2}| = \lim_{n \to ∞} \frac{2(n+1)^2}{5n^2} = \frac{2}{5}< 1$ ⇒ $\sum_{n=1}^\infty \frac{n^2·2^{n-1}}{(-5)^n}$ converges by Radio Test.

As long as we are strictly inside the interval of convergence, we can take derivatives and integrals of power series allowing us to get new series.

Let R be the radius of convergence of a series, x is strictly inside the interval of convergence of the series when −R < x < R.

To differentiate, we simply differentiate each term (not worrying that we have infinitely many terms) and then put the terms back into summation notation. Notice that in the derivative series we must change our index to begin at n = 1.

$\frac{d}{dx}(\sum_{n=0}^\infty a_nx^n) = \frac{d}{dx}(\sum_{n=1}^\infty a_0 + a_1x +a_2x^2 +a_3x^3 +···) = a_1 + 2a_2x + 3a_3x^2 +··· = \sum_{n=1}^\infty na_nx^{n-1}$

$\frac{d}{dx}(\sum_{n=0}^\infty a_n(x-a)^n)=\frac{d}{dx}(\sum_{n=1}^\infty a_0 + a_1(x-a) +a_2(x-a)^2 +a_3(x-a)^3 +···)=a_1+2a_2(x-a)+3a_3(x-a)^2+···=\sum_{n=1}^\infty na_n(x-a)^{n-1}$

$\frac{d}{dx} e^x = \frac{d}{dx}(\sum_{n=0}^\infty \frac{x^n}{n!})=\sum_{n=1}^\infty \frac{(n-1)x^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^\infty \frac{x^m}{m!}=e^x$

$\frac{d}{dx}(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{5^n})=\sum_{n=1}^\infty \frac{(-1)^{n+1}2nx^{2n-1}}{5^n}$

Similarly, while we get an infinite integrand, we don’t worry and just antidifferentiate each term, and then add a constant.

$\int \sum_{n=0}^\infty a_nx^ndx = \int a_0 + a_1x +a_2x^2 +a_3x^3 + ··· dx= c + a_0x + a_1\frac{x^2}{2} + a_2\frac{x^3}{3} + a_3\frac{x^4}{4} + ··· = (\sum_{n=0}^\infty a_n \frac{x^{n+1}}{n+1}) + C$

$\int \sum_{n=0}^\infty a_n(x-a)^n = \int a_0 + a_1(x-a) +a_2(x-a)^2 +a_3(x-a)^3 + ··· dx = c + a_0(x-a) + a_1\frac{(x-a)^2}{2} + a_2\frac{(x-a)^3}{3} + a_3\frac{(x-a)^3}{4} + ··· = (\sum_{n=0}^\infty a_n \frac{(x-a)^{n+1}}{n+1}) + C$

$\int \sum_{n=1}^\infty \frac{x^n}{n!} dx = \sum_{n=1}^\infty \frac{x^{n+1}}{n!·(n+1)}=\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)!}$

$\frac{d}{dx}(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{5^n})=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n+1}}{5^n(2n+1)}$

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].

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