07 Infinite Series

Infinite Series¶

Let \(\set{a_n}_{n \in \mathbb{Z^+}}\) be a sequence. Then \(\sum\limits_{n = 1}^\infty a_n = a_1 + a_2 + \dots\) is called a series.

If the series has a finite number of terms, it is called a finite series; otherwise it is called an infinite series.

A finite series is always convergent.

A infinite series may/ may not be convergent

Series	Type
\(1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} = e^x\)	Converges to \(e^x\)
\(1 + 1 + \dots\)	Divergent
\(1 + \frac12 + \dots + \frac1n\)	Divergent
\(1 - 1 + 1 - 1 + \dots\)	Neither convergent/divergent it is an alternating series which oscilates

If we are able to find the sum of a series, then the series converges to the sum \(S_n = \dfrac{a}{1 - r}\)

if sum is finite, then convergent series
else, divergent series

Series of +ve Terms¶

Consider series \(\sum\limits_{n = 1}^\infty a_n = a_1 + a_2 + \dots + a_n\). This series is a series of +ve terms as \(a_n \ge 0, \forall n\).

We use the following tests.

\(n^\text{th}\) Term Test¶

\[ \lim\limits_{n \to \infty} a_n = \begin{cases} \ne 0 & \text{Divergent} \\ = 0 & \text{Test fails} \end{cases} \]

Important Results¶

Geometric sum \(a + ar + ar^2 + \dots\)
- converges to \(\dfrac{a}{1 - r}, |r| < 1\)
- diverges

		Converges	Diverges
Geometric Series	\(a + ar + ar^2 + \dots\)	\(\vert r \vert < 1\) converges to \(\dfrac{a}{1-r}\)	\(\vert r \vert \ge 1\)
p-series	\(\sum\limits_{n = 1}^\infty \dfrac{1}{n^p}\)	\(p > 1\)	\(p \le 1\)

\[ \begin{aligned} \lim\limits_{n \to \infty} \frac{ \ln \vert n\vert }{n} &= 0 \quad (\ln \vert n \vert \text{ always } < n, \text{ so den reaches } \infty \text{ faster} ) \\ \lim\limits_{n \to \infty} x^{\frac{1}{n}} &= 1 \\ \lim\limits_{n \to \infty} n^{\frac{1}{n}} &= 1 \\ (x^0 = n^0 &= 1) \\ \lim\limits_{n \to \infty} \left( 1 + \frac x n \right)^n &= e^x \\ \lim\limits_{n \to \infty} x^n &= 0 \text{ if } |x| < 1 \\ \lim\limits_{n \to \infty} \frac{x^n}{n!} &= 0 \\ (n! &> x^n), \text{ when } n \text{ is large so den reaches } \infty \text{ faster} \end{aligned} \]

Integral Test¶

This test can be applied when \(a_n = f(n)\) is integrable

Let

\(\sum a_n\) be a series of +ve terms
\(a_n = f(n)\) where \(f\) is
- continuous
- +ve
- decreasing function of \(n\), for some \(n \ge N\)

Then by integral test, \(\int\limits_N^\infty f(x) \ dx\) and \(\sum\limits_N^\infty a_n\) converge/diverge together

\(I\)
Finite	Converges (basically \(S_n\) is finite number)
Infinite	Diverges

Ratio Test¶

Used when series contains factorials like \(n!, (2n)!\)

Let \(\sum a_n\) be a series of +ve terms.

Let \(\lim\limits_{n \to \infty} \dfrac{a_{n+1}}{a_n} = k\)

\(k\)
\(< 1\)	Converges
\(> 1\)	Diverges
\(0, 1\)	Test Fails

Root Test¶

Used when series contains terms with exponents, such as \(n^n, n^{n+1}, n^\frac1n\)

Let \(\lim\limits_{n \to \infty} (a_n)^\frac1n = k\)

\(k\)
\(< 1\)	Converges
\(> 1\)	Diverges
\(1\)	Test Fails

Limit Comparison Test¶

Best used when \(a_n\) is a fraction of polynomial, ie \(a_n = \frac{P(n)}{Q(n)}\), where \(P, Q\) are polynomials in terms of \(n\)

Let

\(\sum a_n\) be a series of +ve terms
\(\sum b_n\) be a known series (we know if it converges/diverges)
- We choose \(b_n = \dfrac{1}{n^{q-p}}\), where
- P = degree of numerator
- Q = degree of denominator
- If \(b_n\) is a p-series of the form \(\sum \dfrac{1}{n^p}\) | \(p\) | | | :-----: | :-------: | | \(> 1\) | converges | | \(\le 1\) | diverges |
- \(\lim\limits_{n \to \infty} \frac{a_n}{b_n} = k\)

Then

Given
\(k = c (\ne 0)\)	both \(\sum a_n\) and \(\sum b_n\) converge
\(k = 0, \sum b_n\) converges	\(\sum a_n\) converges
\(k \to \infty, \sum b_n\) diverges	\(\sum a_n\) diverges

Last Updated: 2024-01-24 ; Contributors: AhmedThahir