where:

• $c_n$ are coefficients (constants)
• $x$ is a variable
• $a$ is the center

💡 Key Idea: A power series is like a polynomial with infinitely many terms!

Power Series Centered at $a = 0$

When $a = 0$, we get a Maclaurin-type power series:

$\sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots$

This is the most common form!

Example 1: Geometric Series as Power Series

The geometric series:

$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$

This is a power series with $c_n = 1$ for all $n$, centered at $a = 0$.

Converges when: $|x| < 1$

Sum: $\frac{1}{1-x}$ (when $|x| < 1$)

Power Series as Functions

A power series defines a function:

$f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$

The domain of $f$ is the set of all $x$ values where the series converges.

Example 2: Recognize the Function

What function does $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ represent?

This is the Taylor series for $e^x$!

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

Converges for all $x$ (we'll prove this later).

Convergence of Power Series

For a given power series $\sum c_n(x-a)^n$, exactly one of the following is true:

1. Converges only at $x = a$ (trivial case)

2. Converges for all $x$ (radius of convergence $R = \infty$)

3. Converges for $|x - a| < R$ and diverges for $|x - a| > R$

where $R$ is the radius of convergence.

The Radius of Convergence

The radius of convergence $R$ is:

• $R = 0$ if series only converges at center
• $0 < R < \infty$ if series converges on an interval
• $R = \infty$ if series converges for all $x$

Interval of convergence: $(a - R, a + R)$ or $[a - R, a + R]$ or variations

(Must check endpoints separately!)

Finding Radius of Convergence

Use the Ratio Test:

$\frac{1}{R} = \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|$

Or equivalently:

$R = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right|$

(If the limit exists or is $\infty$)

Example 3: Find Radius

Find the radius of convergence for $\sum_{n=0}^{\infty} \frac{x^n}{n+1}$.

Use Ratio Test: Let $a_n = \frac{x^n}{n+1}$

$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{\frac{x^{n+1}}{n+2}}{\frac{x^n}{n+1}}\right|$

$= \lim_{n \to \infty} \left|x \cdot \frac{n+1}{n+2}\right|$

$= |x| \cdot \lim_{n \to \infty} \frac{n+1}{n+2} = |x| \cdot 1 = |x|$

For convergence: $|x| < 1$

Radius of convergence: $R = 1$

Example 4: Power Series Centered at $a = 2$

Find the radius of convergence for $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}$.

Use Ratio Test: Let $a_n = \frac{(x-2)^n}{n \cdot 3^n}$

$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x-2)^n}\right|$

$= \lim_{n \to \infty} \left|\frac{x-2}{3} \cdot \frac{n}{n+1}\right|$

$= \frac{|x-2|}{3} \cdot 1 = \frac{|x-2|}{3}$

For convergence: $\frac{|x-2|}{3} < 1$

$|x - 2| < 3$

Radius of convergence: $R = 3$

Center: $a = 2$

Interval (before checking endpoints): $(2-3, 2+3) = (-1, 5)$

Operations on Power Series

If $f(x) = \sum c_n x^n$ and $g(x) = \sum d_n x^n$ both converge for $|x| < R$:

Addition/Subtraction:

$f(x) \pm g(x) = \sum (c_n \pm d_n) x^n$

Scalar Multiplication:

$k \cdot f(x) = \sum (k \cdot c_n) x^n$

Multiplication by $x^m$:

$x^m f(x) = \sum c_n x^{n+m}$

Differentiation of Power Series

Theorem: If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ with radius $R > 0$, then:

$f'(x) = \sum_{n=1}^{\infty} n \cdot c_n (x-a)^{n-1}$

Differentiate term by term!

The derivative has the same radius of convergence $R$.

Example 5: Differentiate Power Series

Find the derivative of $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.

Differentiate term by term:

$f'(x) = \sum_{n=1}^{\infty} \frac{n \cdot x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}$

Reindex with $m = n - 1$:

$= \sum_{m=0}^{\infty} \frac{x^m}{m!}$

Notice: $f'(x) = f(x)$!

This confirms $f(x) = e^x$ (since $\frac{d}{dx}[e^x] = e^x$).

Integration of Power Series

Theorem: If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ with radius $R > 0$, then:

$\int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}$

Integrate term by term!

The integral has the same radius of convergence $R$.

Example 6: Integrate Power Series

Find $\int \sum_{n=0}^{\infty} x^n\,dx$ for $|x| < 1$.

Integrate term by term:

$\int \sum_{n=0}^{\infty} x^n\,dx = C + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$

We know $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x| < 1$.

So: $\int \frac{1}{1-x}\,dx = -\ln|1-x| + C$

This matches: $-\ln(1-x) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{x^n}{n}$

Uniqueness of Power Series

Theorem: If $\sum c_n(x-a)^n = \sum d_n(x-a)^n$ for all $x$ in some interval, then:

$c_n = d_n \text{ for all } n$

Power series representation is unique!

Example 7: Find Coefficients

If $\sum_{n=0}^{\infty} c_n x^n = \frac{1}{1-x}$ for $|x| < 1$, find $c_n$.

We know: $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

By uniqueness: $c_n = 1$ for all $n \geq 0$.

Substitution in Power Series

Can substitute $u = f(x)$ into a power series!

Example: If $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x| < 1$:

Replace $x$ with $-x^2$:

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

$1 - x^2 + x^4 - x^6 + \cdots = \frac{1}{1+x^2}$ (for $|x| < 1$)

⚠️ Common Mistakes

Mistake 1: Forgetting Absolute Value

WRONG: $|x - 2| < 3$ means $x < 5$

RIGHT: $|x - 2| < 3$ means $-3 < x - 2 < 3$, so $-1 < x < 5$

Mistake 2: Not Checking Endpoints

Radius tells you interval is $(a-R, a+R)$, but series might converge at one or both endpoints!

Must check $x = a - R$ and $x = a + R$ separately.

Mistake 3: Wrong Index After Differentiation

When differentiating $\sum_{n=0}^{\infty} c_n x^n$:

Get $\sum_{n=1}^{\infty} n c_n x^{n-1}$ (starts at $n=1$, not $n=0$!)

The $n=0$ term disappears (constant → 0).

Mistake 4: Confusing Center and Radius

For $\sum c_n(x-2)^n$ with $R = 3$:

• Center: $a = 2$
• Radius: $R = 3$
• Interval: $(-1, 5)$ before checking endpoints

Not centered at 0!

📝 Practice Strategy

1. Identify center $a$: Look at $(x - a)^n$
2. Use Ratio Test: Find $\lim|\frac{a_{n+1}}{a_n}|$, set $< 1$ for convergence
3. Solve for $x$: Get $|x - a| < R$
4. Check endpoints: Test $x = a \pm R$ separately using other tests
5. For differentiation: Differentiate each term, start sum at $n=1$
6. For integration: Integrate each term, add constant
7. Known series: Memorize $\frac{1}{1-x} = \sum x^n$ and $e^x=\sum \frac{x^{}}{}$

for $|x| < 1$.

Step 2: Integrate both sides

$\int \frac{1}{1+x}\,dx = \int \sum_{n=0}^{\infty} (-1)^n x^n\,dx$

$\ln|1+x| + C = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$

Step 3: Find constant

At $x = 0$: $\ln 1 = 0$, and the series sum is 0.

So $C = 0$.

For $-1 < x < 1$: $1 + x > 0$, so $|1+x| = 1+x$.

$\ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$

Step 4: Reindex

Let $m = n + 1$:

$\ln(1+x) = \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}$

Or more commonly written as:

$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n$

$= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

for $|x| < 1$.

Answer: $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n$ for $|x| < 1$

Step 2: Find first derivative

$f'(x) = \sum_{n=1}^{\infty} \frac{2n \cdot x^{2n-1}}{(2n)!}$

$= \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!}$

$= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$

Step 3: Find second derivative

$f''(x) = \sum_{n=1}^{\infty} \frac{(2n-1) \cdot x^{2n-2}}{(2n-1)!}$

$= \sum_{n=1}^{\infty} \frac{x^{2n-2}}{(2n-2)!}$

Reindex with $m = n - 1$:

$= \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m)!} = f(x)$

Step 4: Identify the function

We have $f''(x) = f(x)$ with $f(0) = 1$ and $f'(0) = 0$.

This is the differential equation for $\cos x$!

But wait, $f(0) = \cos 0 = 1$ ✓

And $f'(0) = -\sin 0 = 0$ ✓

Actually, $\frac{d^2}{dx^2}[\cos x] = -\cos x$, not $+\cos x$.

Let me reconsider...

If $g(x) = \cosh x = \frac{e^x + e^{-x}}{2}$, then:

$\cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$

And $\frac{d^2}{dx^2}[\cosh x] = \cosh x$ ✓

Answer: $f''(x) = f(x)$, and $f(x) = \cosh x$ (hyperbolic cosine)

Or: $f(x) = \frac{e^x + e^{-x}}{2}$