Summation of Algebraic & Trigonometric Series

1. The Axiom of Mathematical Induction

The Principle of Mathematical Induction is fundamentally equivalent to the Well-Ordering Principle of the natural numbers $\mathbb{N}$:

• Weak Induction: If a proposition $P(n)$ is true for $n = 1$, and for every $k \ge 1$ the truth of $P(k)$ implies $P(k+1)$, then $P(n)$ is true for all $n \in \mathbb{N}$.
• Strong Induction: If $P(1)$ is true, and the truth of $P(1), P(2), \dots, P(k)$ collectively implies $P(k+1)$, then $P(n)$ is true for all $n \in \mathbb{N}$.

2. Canonical Power Sum Identities

Inductive proofs establish the classic closed forms for integer power sums:

1. The Difference Operator $\Delta$

For a sequence $f(n)$, the forward difference operator is defined by: $$\mathbf{\Delta f(n) \equiv f(n+1) - f(n)}$$ The Fundamental Theorem of Summation Calculus states that the sum of differences telescopes: $$\mathbf{\sum_{k=1}^n \Delta f(k) = \sum_{k=1}^n [f(k+1) - f(k)] = f(n+1) - f(1)}$$ All intermediate terms cancel pairwise, leaving only the boundary evaluations!

2. Factorial Polynomials

We define the falling factorial polynomial of degree $r$: $$n^{(r)} \equiv n(n-1)(n-2)\cdots(n-r+1)$$ Its forward difference mirrors standard polynomial differentiation: $$\Delta [n^{(r)}] = (n+1)^{(r)} - n^{(r)} = r \, n^{(r-1)}$$ Thus, summation follows the discrete power rule: $$\mathbf{\sum_{k=1}^n k^{(r)} = \frac{(n+1)^{(r+1)} - 1^{(r+1)}}{r + 1} = \frac{(n+1)^{(r+1)}}{r + 1}}$$ Any polynomial can be converted to factorial powers via Stirling numbers of the second kind, rendering summation algorithmic!

1. Definition of an AGP

An Arithmetico-Geometric Progression is a sequence whose $k$-th term is the product of corresponding terms of an Arithmetic Progression ($a, a+d, a+2d, \dots$) and a Geometric Progression ($1, r, r^2, \dots$): $$\mathbf{u_k = [a + (k-1)d] r^{k-1}}$$ The finite sum to $n$ terms is: $$S_n = a + (a + d)r + (a + 2d)r^2 + \dots + [a + (n-1)d]r^{n-1}$$

2. Closed-Form Derivation

Multiply $S_n$ by the common ratio $r$: $$r S_n = ar + (a + d)r^2 + \dots + [a + (n-2)d]r^{n-1} + [a + (n-1)d]r^n$$ Subtracting this equation from $S_n$: $$(1 - r)S_n = a + d[r + r^2 + \dots + r^{n-1}] - [a + (n-1)d]r^n$$ The bracketed terms form a standard finite geometric series with sum $\frac{r(1 - r^{n-1})}{1 - r}$: $$(1 - r)S_n = a + \frac{dr(1 - r^{n-1})}{1 - r} - [a + (n-1)d]r^n$$ Dividing by $(1 - r)$ yields the exact closed form: $$\mathbf{S_n = \frac{a}{1 - r} + \frac{dr(1 - r^{n-1})}{(1 - r)^2} - \frac{[a + (n-1)d]r^n}{1 - r}}$$

3. Sum to Infinity

For $|r| < 1$, as $n \to \infty$, $r^n \to 0$ and $n r^n \to 0$. The infinite sum simplifies to: $$\mathbf{S_\infty = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2}}$$

1. The Method of Partial Fractions for Series

When terms of an infinite series are reciprocal products of linear factors, we decompose each term $u_k$ into partial fractions to induce telescoping cancellation: $$u_k = \frac{1}{(k + a)(k + b)} = \frac{1}{b - a} \left[ \frac{1}{k + a} - \frac{1}{k + b} \right]$$ Summing from $k = 1$ to $n$: $$S_n = \frac{1}{b - a} \sum_{k=1}^n \left( \frac{1}{k + a} - \frac{1}{k + b} \right)$$ Depending on the shift $b - a$, intermediate terms cancel, leaving a finite number of uncancelled boundary fractions.

2. Higher-Order Factor Decompositions

For three factors in arithmetic progression: $$u_k = \frac{1}{(a k + b)(a(k+1) + b)(a(k+2) + b)} = \frac{1}{2a} \left[ \frac{1}{(ak+b)(a(k+1)+b)} - \frac{1}{(a(k+1)+b)(a(k+2)+b)} \right]$$ Defining $v_k = \frac{1}{(ak+b)(a(k+1)+b)}$, we observe that $u_k = \frac{1}{2a}[v_k - v_{k+1}]$. The sum to $n$ terms telescopes directly: $$\sum_{k=1}^n u_k = \frac{1}{2a}[v_1 - v_{n+1}]$$ and as $n \to \infty$, $v_{n+1} \to 0$, giving the exact limit $S_\infty = \frac{v_1}{2a}$!

1. The Complex Phasor Coupling Technique

To sum a cosine series $C = \sum_{k=0}^{n-1} a_k \cos(\theta_k)$ and sine series $S = \sum_{k=0}^{n-1} a_k \sin(\theta_k)$, we form the complex linear combination: $$\mathbf{C + iS = \sum_{k=0}^{n-1} a_k [\cos(\theta_k) + i\sin(\theta_k)] = \sum_{k=0}^{n-1} a_k e^{i\theta_k}}$$ This converts trigonometric sums into geometric or binomial series in the complex domain! After evaluating $C + iS$ in closed form, equating real parts recovers $C$, and equating imaginary parts recovers $S$.

2. Sum of Sines and Cosines in Arithmetic Progression

Let $\theta_k = \alpha + k\beta$: $$C + iS = \sum_{k=0}^{n-1} e^{i(\alpha + k\beta)} = e^{i\alpha} \sum_{k=0}^{n-1} (e^{i\beta})^k = e^{i\alpha} \frac{1 - e^{in\beta}}{1 - e^{i\beta}}$$ Factoring out half-angles: $$1 - e^{in\beta} = -e^{in\beta/2}(e^{in\beta/2} - e^{-in\beta/2}) = -2i e^{in\beta/2}\sin(n\beta/2)$$ $$1 - e^{i\beta} = -2i e^{i\beta/2}\sin(\beta/2)$$ Dividing: $$C + iS = e^{i\alpha} \frac{-2i e^{in\beta/2}\sin(n\beta/2)}{-2i e^{i\beta/2}\sin(\beta/2)} = \frac{\sin(n\beta/2)}{\sin(\beta/2)} e^{i\left(\alpha + \frac{n-1}{2}\beta\right)}$$ Equating real and imaginary parts gives the famous closed forms: