Power Series. Convergence, derivatives, and Integrals

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

Infinite Series. Convergence and Divergence.

An infinite series is the sum of an infinite sequence of numbers. It is represented as: $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \dots$ 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 finite limit, that is, $l = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^n a_k$ exists and is a finite number.

The $n$-th partial sum $S_n$ is the sum of the first $n$ terms: $S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n$. If the limit $L = \lim_{n \to \infty} S_n$ exists and is a finite number, the series converges, and $L$ is called the sum of the series. Formally, this means for any arbitrarily small positive number $\epsilon$, there is a sufficiently large $N$ such that for all $n \geq N$, $\vert{}S_n - L\vert{} < \epsilon$.

If this limit does not exist (e.g., it goes to infinity or oscillates forever), the series is said to be divergent.

The Divergence Test (n-th Term Test). If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum_{n=1}^\infty a_n$ 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 $\sum a_n$ and $\sum b_n$ be series with positive terms such that $0 \leq a_n \leq b_n$ for all $n \geq N$.

1. If the “bigger” series $\sum b_n$ converges, then the “smaller” series $\sum a_n$ converges.
2. If the “smaller” series $\sum a_n$ diverges, then the “bigger” series $\sum b_n$ diverges.
3. If the bigger series diverges, or the smaller series converges, the test is inconclusive.

Power Series

A geometric series is the sum of an infinite number of terms that have a constant ratio between successive terms. It is written as $a + ar + ar^2 + ar^3 + \dots$, 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 is: $\sum_{n=0}^\infty ar^n = \frac{a}{1-r} \quad \text{if } \vert{}r\vert{} < 1$ and that the series diverges when $\vert{}r\vert{} \ge 1$. We typically think of $r$ as a constant, but we can also treat it as a variable.

In this case, the series becomes a function of $x$: $\boxed{f(x) = \sum_{n=0}^\infty ax^n = \frac{a}{1-x} \quad \text{for } \vert{}x\vert{} < 1}$

Definition: Power Series

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

Power Series Convergence

Because the terms in a power series involve a variable $x$, the series may converge for certain values of $x$ and diverge for others. A power series centered at $x = a$ always converges at its center (when $x=a$, all terms except $a_0$ are zero, so the sum is simply $a_0$). Most power series, however, converge for more than just one value of $x$.

Theorem: Convergence of a Power Series

Consider the power series $\sum_{n=0}^\infty a_n(x-a)^n$. It satisfies exactly one of the following three properties:

1. The series converges only at the center $x = a$ and diverges for all $x \neq a$.
2. The series converges absolutely for all real numbers $x \in \mathbb{R}$.
3. There exists a real number $R > 0$ (called the radius of convergence) such that the series converges absolutely for all $x$ where $\vert{}x-a\vert{} < R$, and diverges for all $x$ where $\vert{}x-a\vert{} > R$.

Proof.

For the sake of simplicity, assume the power series is centered at $a = 0$. (If it isn’t, we can just use the substitution $y = x - a$ and analyze $\sum_{n=0}^\infty a_ny^n$).

Part 1: Proving convergence inside a bounded radius

1. Suppose there exists a point $d \neq 0$ such that $\sum_{n=0}^\infty a_nd^n$ converges. We need to prove that $\sum_{n=0}^\infty a_nx^n$ converges absolutely for all $x$ where $\vert{}x\vert{} < \vert{}d\vert{}$.
2. Because $\sum_{n=0}^\infty a_nd^n$ converges, the $n$-th term test for divergence tells us its terms must approach zero: $\lim_{n \to \infty} a_nd^n = 0$.
3. This implies the terms are bounded. There exists an integer $N$ such that $\vert{}a_nd^n\vert{} \le 1$ for all $n \ge N$.
4. Let’s rewrite the term we are testing: $\vert{}a_nx^n\vert{} = \left\vert{}a_nd^n \left(\frac{x}{d}\right)^n\right\vert{} = \vert{}a_nd^n\vert{} \left\vert{}\frac{x}{d}\right\vert{}^n$
5. Using our bound from step 3, for all $n \ge N$: $\vert{}a_nx^n\vert{} \le 1 \cdot \left\vert{}\frac{x}{d}\right\vert{}^n$
6. Now consider the series $\sum_{n=N}^\infty \left\vert{}\frac{x}{d}\right\vert{}^n$. This is a geometric series with common ratio $r = \left\vert{}\frac{x}{d}\right\vert{}$. Because we stated that $\vert{}x\vert{} < \vert{}d\vert{}$, the ratio $r < 1$, which means this geometric series converges.
7. By the Direct Comparison Test, since $\sum_{n=N}^\infty \left\vert{}\frac{x}{d}\right\vert{}^n$ converges, the series $\sum_{n=N}^\infty \vert{}a_nx^n\vert{}$ also converges. Adding back the finite number of missing initial terms does not affect infinite convergence. Thus, the series converges absolutely for all $\vert{}x\vert{} < \vert{}d\vert{}$.

Part 2: Defining the Radius $R$

1. Let $S$ be the set of real numbers for which the series $\sum_{n=0}^\infty a_nx^n$ converges.
2. If $S = \{0\}$, we are in Case 1.
3. If $S = \mathbb{R}$, we are in Case 2.
4. Otherwise, if $S \neq \{0\}$ and $S \neq \mathbb{R}$, there exists a real number $s \neq 0$ where the series diverges. Based on what we just proved, the series cannot converge for any $x$ where $\vert{}x\vert{} > \vert{}s\vert{}$ (if it did, it would have to converge at $s$, which is a contradiction).
   Therefore, $S$ must be a bounded set.
   By the Least Upper Bound Property of real numbers, this set has a supremum (smallest upper bound).
   We call this supremum $R$. The series converges for $\vert{}x\vert{} < R$ and diverges for $\vert{}x\vert{} > R$. This proves Case 3. $\blacksquare$

The Least Upper Bound Property states that every non-empty set of real numbers with an upper bound has a least upper bound (or supremum) that is also a real number. This fundamental rule means the real number line has no gaps or empty space

Solved Examples

Ratio Test & Endpoints

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

1. Apply the Ratio Test
   $L = \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{\frac{(-1)^{n+1}x^{n+1}}{n+2}}{\frac{(-1)^nx^n}{n+1}} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{x^{n+1}}{n+2} \cdot \frac{n+1}{x^n} \right\vert{} = \lim_{n \to \infty} \vert{}x\vert{} \left\vert{} \frac{n+1}{n+2} \right\vert{} = \vert{}x\vert{}$
2. For the series to converge, we need $L < 1$, so $\vert{}x\vert{} < 1$. This means the radius of convergence is $R = 1$.
3. Check the Endpoints ($x = 1, x = -1$).
   Let $x = -1$: $\sum_{n=0}^\infty \frac{(-1)^n(-1)^n}{n+1} = \sum_{n=0}^\infty \frac{1}{n+1} = \sum_{n=1}^\infty \frac{1}{n}$. This is the Harmonic Series, which diverges. So, $x = -1 \notin I$.
   Let $x = 1$: $\sum_{n=0}^\infty \frac{(-1)^n(1)^n}{n+1} = \sum_{n=0}^\infty \frac{(-1)^n}{n+1}$. This is the Alternating Harmonic Series. By the Alternating Series Test, because $\frac{1}{n+1}$ decreases monotonically to $0$ (vanishing limit), it converges. So, $x = 1 \in I$.
4. Conclusion: The interval of convergence is $I = (-1, 1]$.

Alternating Series Test. An alternating series $\sum_{n=0}^\infty (-1)^nb_n$ where $b_n = \left\vert{} a_n \right\vert{}$ converges if decreases monotonically ($b_{n+1} \leq b_n$ eventually) and vanishing limit ($\lim_{n \to ∞} b_n = 0$).

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

1. Let’s use the Ratio Test: $\lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{\frac{(n+1)^2 \cdot 2^{n}}{(-5)^{n+1}}}{\frac{n^2 \cdot 2^{n-1}}{(-5)^n}} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{(n+1)^2 \cdot 2^n}{(-5)^{n+1}} \cdot \frac{(-5)^n}{n^2 \cdot 2^{n-1}} \right\vert{}$
   Simplify the exponents ($2^n / 2^{n-1} = 2$ and $(-5)^n / (-5)^{n+1} = -1/5$): $= \lim_{n \to \infty} \left\vert{} \frac{2(n+1)^2}{-5n^2} \right\vert{} = \frac{2}{5} \lim_{n \to \infty} \frac{n^2+2n+1}{n^2} = \frac{2}{5} \cdot 1 = \frac{2}{5}$
2. Because $\frac{2}{5} < 1$, the series converges by the Ratio Test.

General Rule for the Ratio Test

In general, for an arbitrary power series $\sum_{n=0}^\infty a_nx^n$: $\lim_{n \to \infty} \left\vert{} \frac{a_{n+1}x^{n+1}}{a_nx^n} \right\vert{} = \lim_{n \to \infty} \vert{}x\vert{}\frac{\vert{}a_{n+1}\vert{}}{\vert{}a_n\vert{}} = L\vert{}x\vert{}$

Assuming $L = \lim_{n \to \infty} \frac{\vert{}a_{n+1}\vert{}}{\vert{}a_n\vert{}}$ exists:

1. If $L \in (0, \infty)$, the series converges if $L\vert{}x\vert{} < 1 \iff \vert{}x\vert{} < \frac{1}{L}$. The radius of convergence is $R = \frac{1}{L}$. Endpoints $x = \pm \frac{1}{L}$ require manual checking.
2. If $L = 0$, then $L\vert{}x\vert{} = 0$ for all real $x$. The series converges for all $x$, so $R = \infty$ and $I = (-\infty, \infty)$.
3. If $L = \infty$, the limit exceeds 1 for all $x \neq 0$. The series converges only at its center, so $R = 0$.

Using the Root Test

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

1. ⚠️ If $n=1$, the denominator becomes $(1-1)^1 = 0$, causing division by zero. That’s why we need to adjust the index to start at $n=2$.
2. Apply the Root Test: $\lim_{n \to \infty} \vert{}a_n\vert{}^{\frac{1}{n}} = \lim_{n \to \infty} \left\vert{} \frac{(-1)^n \cdot 3^{n+2}}{(n-1)^n} \right\vert{}^{\frac{1}{n}} = \lim_{n \to \infty} \left( \frac{3^2 \cdot 3^n}{(n-1)^n} \right)^{\frac{1}{n}}$
   $= \lim_{n \to \infty} \frac{9^{\frac{1}{n}} \cdot 3}{n-1} = \frac{1 \cdot 3}{\infty} = 0$
   Because $0 < 1$, the series converges absolutely by the Root Test.

Constructing Power Representations

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

1. Recall the geometric series template: $\sum_{n=0}^\infty u^n = \frac{1}{1-u}$ for $\vert{}u\vert{} < 1$.
2. We can rewrite $f(x)$ to match this template by letting $u = x^4$: $f(x) = 3 \cdot \frac{1}{1-(x^4)} = 3 \sum_{n=0}^\infty (x^4)^n = \sum_{n=0}^\infty 3x^{4n}$
3. Convergence requires $\vert{}x^4\vert{} < 1 \implies x^4 < 1 \implies -1 < x < 1$.
   $x^4 -1 \lt 0 \implies (x^2-1)(x^2+1) \implies[x^2+1 \gt 0, \forall x] x^2 -1 \lt 0 \implies x^2 \lt 1 \implies -1 < x < 1$
4. The interval of convergence is $I = (-1, 1)$.

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

1. Rewrite $f(x)$ to match the geometric template: $f(x) = x \cdot \frac{1}{1 - (-2x^2)}$
   Let $u = -2x^2$: $f(x) = x \sum_{n=0}^\infty (-2x^2)^n = x \sum_{n=0}^\infty (-1)^n 2^n x^{2n} = \sum_{n=0}^\infty (-1)^n 2^n x^{2n+1}$
2. Convergence requires $\vert{}-2x^2\vert{} < 1 \implies 2x^2 < 1 \implies x^2 < \frac{1}{2} \implies \vert{}x\vert{} < \frac{1}{\sqrt{2}}$.
3. The interval of convergence is $I = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

Derivatives and Integrals of Power Series

One of the most powerful features of power series is that, as long as we are strictly inside their interval of convergence, we can treat them like infinitely long polynomials. This allows us to take derivatives and integrals term-by-term to generate entirely new series.

Theorem. Term-by-Term Differentiation and Integration. If a power series $\sum a_n(x-a)^n$ has a radius of convergence $R > 0$, then the function $f(x)$ defined by the series is differentiable (and therefore continuous and integrable) on the interval $(a-R, a+R)$.

Notice that when you differentiate or integrate a power series, the radius of convergence $R$ remains exactly the same. However, convergence at the endpoints ($x = a-R$ and $x = a+R$) can change and must be re-tested for the new series.

Term-by-Term Differentiation

To differentiate a power series, we simply take the derivative of each term individually (using the power rule) and then put the terms back into summation notation.

Notice that in the derivative series, the constant term $a_0$ disappears, so we shift our starting index from $n = 0$ to $n = 1$.

General Formulas

• Centered at $x = 0$: $\frac{d}{dx} \left( \sum_{n=0}^\infty a_nx^n \right) = \frac{d}{dx} \left( a_0 + a_1x + a_2x^2 + a_3x^3 + \dots \right) = a_1 + 2a_2x + 3a_3x^2 + \dots = \sum_{n=1}^\infty na_nx^{n-1}$
• Centered at $x = a$: $\frac{d}{dx} \left( \sum_{n=0}^\infty a_n(x-a)^n \right) = \frac{d}{dx} \left( a_0 + a_1(x-a) + a_2(x-a)^2 + a_3(x-a)^3 + \dots \right)= a_1 + 2a_2(x-a) + 3a_3(x-a)^2 + \dots = \sum_{n=1}^\infty na_n(x-a)^{n-1}$

Solved Examples

• The derivative of $e^x$

Let’s verify that the derivative of $e^x$ is itself using its Maclaurin series: $\frac{d}{dx} e^x = \frac{d}{dx} \left( \sum_{n=0}^\infty \frac{x^n}{n!} \right) = \sum_{n=1}^\infty \frac{n x^{n-1}}{n!}$

Since $n! = n \cdot (n-1)!$, the $n$ in the numerator and denominator cancel: $= \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$

To make this look like our original series, we can re-index by letting $m = n - 1$. When $n=1$, $m=0$: $\frac{d}{dx} e^x = \sum_{m=0}^\infty \frac{x^m}{m!} = e^x$

• Differentiating a more complex series $ \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{5^n}$

1. The series converges for all 𝑥. It’s a geometric-type series $\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{5^n} = \sum_{n=1}^\infty (-1)^{n+1}(\frac{x^2}{5})^n$ with ratio $r = -\frac{x^2}{5}$.
2. For convergence we need $|r| = |\frac{x^2}{5}| \lt 1 \implies x^2 \lt 5 \implies |x| \lt \sqrt{5}$. So the radius of convergence is $R = \sqrt{5}$
3. Therefore, we may differentiate term‑by‑term: $\frac{d}{dx} \left( \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{5^n} \right) = \left( \sum_{n=1}^\infty \frac{(-1)^{n+1}}{5^n} \frac{d}{dx} (x^{2n}) \right)$
4. Apply the power rule to $x^{2n}$ (pull down $2n$ and subtract 1 from the exponent):$= \sum_{n=1}^\infty \frac{(-1)^{n+1}(2n)x^{2n-1}}{5^n}$

Term-by-Term Integration

Similarly, even though we have an infinite integrand, we don’t have to worry. We just find the antiderivative of each term individually using the reverse power rule, and then add a single constant of integration $C$ to the final result.

General Formulas

• Centered at $x = 0$: $\int \left( \sum_{n=0}^\infty a_nx^n \right) dx = \int (a_0 + a_1x + a_2x^2 + a_3x^3 + \dots) dx = C + a_0x + a_1\frac{x^2}{2} + a_2\frac{x^3}{3} + a_3\frac{x^4}{4} + \dots = C + \sum_{n=0}^\infty a_n \frac{x^{n+1}}{n+1}$
• Centered at $x = a$: $\int \left( \sum_{n=0}^\infty a_n(x-a)^n \right) dx = \int (a_0 + a_1(x-a) + a_2(x-a)^2 + a_3(x-a)^3 + \dots) dx = C + a_0(x-a) + a_1\frac{(x-a)^2}{2} + a_2\frac{(x-a)^3}{3} + \dots = C + \sum_{n=0}^\infty a_n \frac{(x-a)^{n+1}}{n+1}$

Solved Examples

• Let’s say we have the series expansion for $e^x - 1$, which starts at $n=1$: $\int \left( \sum_{n=1}^\infty \frac{x^n}{n!} \right) dx = C + \sum_{n=1}^\infty \frac{x^{n+1}}{n!(n+1)}$

Since $n! \cdot (n+1) = (n+1)!$, this simplifies nicely to: $= C + \sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)!}$

• Integrating an alternating series. $\int \left( \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n}}{5^n} \right) dx$

Apply the reverse power rule to $x^{2n}$ (add 1 to the exponent and divide by the new exponent): $= C + \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n+1}}{5^n(2n+1)}$

Bibliography

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.