July 06, 2026 - Calculus 2 - Power Series and Convergence

Introduction to Power Series

• A power series is structured as an infinite polynomial. Unlike a standard polynomial which has a finite degree, the degree of a power series is infinite because the powers of the variable $x$ increase indefinitely.

• The general form of a power series is often written using the terms of a sequence $C_n$:     $C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \dots + C_n x^n + \dots$

• In this sequence, the notation starts at indices of zero:

  • The first term is $C_0$.

  • The power of $x$ also starts at zero, where $x^0 = 1$.

  • Each term in the sequence $C_n$ is paired with a corresponding power of $x$ ($x^n$).

• An example of a basic power series is the series where all coefficients are 1:     $\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots$

  • If we plug in $n=0$, we get $1$.

  • If we plug in $n=1$, we get $x$.

  • The sequence of coefficients for this series is simply a constant series of $1$.

Using the Ratio Test for Convergence

• The ratio test is a primary tool used to determine the convergence of power series.

• The test involves finding the limit as $n$ approaches infinity of the absolute value of the ratio of the $(n+1)$-th term to the $n$-th term:     $\rho = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$

• Applying this to the geometric series $\sum x^n$:

  • $a_{n+1} = x^{n+1}$

  • $a_{n} = x^n$

  • The ratio is $\frac{x^{n+1}}{x^n} = x$.

• Convergence Criteria: The ratio test states that the series converges if the result of the limit $\rho$ is strictly less than 1 (\rho < 1).

• For the geometric series, this means |x| < 1. This inequality implies that $x$ must be in the interval -1 < x < 1.

• Within this specific interval, the infinite series represents a defined function:     $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

• Outside of this interval (where $|x| \ge 1$), the series diverges and does not represent the function. For example, at $x=1$, the series is just adding $1$ repeatedly, and at $x=-1$, it oscillates between $1$ and $0$.

Shifting the Center of a Power Series

• A series that is purely in terms of $x^n$ is said to be centered at $x = 0$. This is because if you plug in $x=0$, the series converges instantly as you are simply adding up zeros (after the initial constant term).

• To shift the center of the series, we replace $x$ with $x - a$. This results in a series centered at $x = a$:     $\sum_{n=0}^{\infty} C_n (x - a)^n$

• Example: A series centered at $x=5$ would look like $\sum (x - 5)^n$:     $(x-5)^0 + (x-5)^1 + (x-5)^2 + \dots$

• Determining convergence for shifted series using the ratio test:     $\lim_{n \to \infty} \left| \frac{(x-5)^{n+1}}{(x-5)^n} \right| = |x-5|$

  • Convergence requires |x-5| < 1.

  • This removes the absolute value to become -1 < x-5 < 1.

  • Adding $5$ to all parts yields 4 < x < 6.

  • In the middle of this interval is the center, $5$. The function this series represents would be $\frac{1}{1-(x-5)}$, which simplifies to $\frac{1}{6-x}$, but only on that interval.

Analyzing Specific Case Studies: e^x and n!x^n

• The Exponential Function $\mathbf{e^x}$:

  • The power series for $e^x$ involves factorials: $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

  • Writing out terms: $\frac{1}{0!} + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$

  • Ratio Test: $\lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1)!} \times \frac{n!}{x^n} \right| = \lim_{n \to \infty} \left| \frac{x}{n+1} \right|$.

  • Since $x$ is held constant while the denominator $n+1$ goes to infinity, the limit is $0$.

  • $0$ is always less than $1$, meaning this series converges everywhere, for any value of $x$. Its interval of convergence is $(-\infty, \infty)$.

• The Factorial Series $\mathbf{n!x^n}$:

  • Each term in this series grows significantly larger than the previous one.

  • Ratio Test: $\lim_{n \to \infty} \left| \frac{(n+1)! x^{n+1}}{n! x^n} \right| = \lim_{n \to \infty} |(n+1)x|$.

  • As $n \to \infty$, this limit becomes infinity (unless $x=0$).

  • Infinity is never less than $1$, so this series diverges everywhere except at its center, where $x=0$. This is considered a non-interesting or trivial convergence.

The Radius of Convergence (R)

• Any power series will exhibit one of three behaviors regarding its convergence:

  1. Converges on an Interval: There is a center $a$ and a radius $R$ such that it converges for |x-a| < R. This is an interval from $a-R$ to $a+R$.

  2. Converges Everywhere: The radius $R = \infty$. The interval is $(-\infty, \infty)$.

  3. Converges Only at the Center: The radius $R = 0$. It only converges at $x=a$.

• The radius is "half the length" of the total interval. For the geometric series centered at $0$ on $(-1, 1)$, the total width is $2$, so the radius $R=1$.

Determining Convergence at Endpoints

• The ratio test determines the "open" interval of convergence, but it does not clarify what happens exactly at the boundaries where the ratio equals 1. These points must be checked individually.

• Example calculation for $\sum_{n=0}^{\infty} \frac{(x-2)^n}{n+1} \times \frac{1}{3^n}$:

  • Ratio test: $\lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{(n+2) 3^{n+1}} \times \frac{(n+1) 3^n}{(x-2)^n} \right| = \lim_{n \to \infty} \left| \frac{(x-2) (n+1)}{3(n+2)} \right| = \frac{|x-2|}{3}$.

  • Setting \frac{|x-2|}{3} < 1 \implies |x-2| < 3 \implies -3 < x-2 < 3.

  • This gives the interval -1 < x < 5. The radius is $R=3$.

• Testing Boundary $x=-1$:

  • The series becomes $\sum \frac{(-1-2)^n}{(n+1)3^n} = \sum \frac{(-3)^n}{(n+1)3^n} = \sum \frac{(-1)^n}{n+1}$.

  • Using the Alternating Series Test:

    1. The limit of terms is $0$: $\lim_{n \to \infty} \frac{1}{n+1} = 0$.

    2. The sequence is decreasing: \frac{1}{n+2} < \frac{1}{n+1}.

  • Conclusion: The series converges at $x=-1$.

• Testing Boundary $x=5$:

  • The series becomes $\sum \frac{(5-2)^n}{(n+1)3^n} = \sum \frac{3^n}{(n+1)3^n} = \sum \frac{1}{n+1}$.

  • This is the harmonic series (a P-series with $p=1$), which is known to diverge.

• Final Interval of Convergence: $[-1, 5)$.

Algebraic Properties and Manipulations

If two power series converge to functions $f(x)$ and $g(x)$ on a common interval $I$, the following operations are valid:

1. Addition/Subtraction: $\sum (C_n \pm d_n)x^n = f(x) \pm g(x)$.

2. Scalar/Variable Multiplication: A series can be multiplied by a constant $m$ or a power of $x$ ($x^k$) by distributing it inside the summation: $\sum m C_n x^{n+k} = m x^k f(x)$.

3. Composition (Substitution): Replacing $x$ in a known series with another expression such as $bx^m$. For example, if we have the series for $\frac{1}{1-x}$, replacing $x$ with $x^2$ yields $\sum (x^2)^n = \sum x^{2n}$, which converges to $\frac{1}{1-x^2}$.

Complex Example: Partial Fractions to Power Series

• To find the power series for a rational function like $f(x) = \frac{1}{x^2 - 4x + 3}$, first use partial fractions.

• Factoring the denominator gives $(x-3)(x-1)$. Partial fraction decomposition yields:     $\frac{1}{(x-1)(x-3)} = \frac{1/2}{x-3} - \frac{1/2}{x-1}$

• Transform these into the form $\frac{1}{1-u}$:

  • For the second term: $\frac{-1/2}{x-1} = \frac{1/2}{1-x} = \frac{1}{2} \sum x^n$.

  • For the first term: $\frac{1/2}{x-3} = \frac{1/2}{-3(1 - x/3)} = \frac{-1}{6} \sum (x/3)^n$.

• The resulting combined series is:     $\sum_{n=0}^{\infty} \left( \frac{1}{2} x^n - \frac{1}{6} \left(\frac{x}{3}\right)^n \right) = \sum_{n=0}^{\infty} \frac{1}{2} \left( 1 - \frac{1}{3^{n+1}} \right) x^n$

• Determining the radius for the sum:

  • The first piece converges on $(-1, 1)$.

  • The second piece converges on $(-3, 3)$.

  • The entire sum converges only on the overlap of these intervals, which is $(-1, 1)$. Thus, the radius of convergence for the sum is $R=1$.

Preview of Trigonometric Series

• Power series can be used to represent trigonometric functions.

• Sine Function ($\mathbf{\sin(x)}$): Represented by an alternating series of all odd powers:     $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

• Cosine Function ($\mathbf{\cos(x)}$): Represented by an alternating series of all even powers:     $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$

• Both of these series involve factorials and converge everywhere ($R = \infty$), effectively defining the functions through their infinite growth.

Questions & Discussion

• Question regarding "Row" vs "P": A student asked if the symbol used in the ratio test is the Greek letter $\rho$ (rho) or $P$.

• Response: The instructor clarified that it is $\rho$ (rho). While some might mistake it for $P$, in calculus, $P$ is usually reserved for the "P-series" specifically referring to the power in the denominator ($\sum \frac{1}{n^p}$). For the ratio test, any symbol works, but $\rho$ is conventional.

• Question regarding endpoints on Geometric Series: A student asked if endpoints $-1$ and $1$ are typically included.

• Response: They are not included in the geometric series because at $1$, the terms are $1+1+1...$ (divergent) and at $-1$, it is $1-1+1-1...$ (divergent). You must always check individual series to see if terms decrease to zero at the boundaries.

• Question regarding the parity of Sine: Is sine an even or odd number?

• Response: Sine is an odd function, which matches the fact that its power series consists only of odd powers of $x$.