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Mathematical induction

Prove formulas for every finite payment count using a base case and a general inductive step, and distinguish induction from numerical checks and infinite limits.

Mathematical induction is a proof method for statements indexed by integers: prove a starting case, then prove that each case implies the next. In quantitative work, it can certify a formula for every finite payment count or establish a property preserved by every iteration of an algorithm. Checking a few outputs can reveal a mistake; induction can establish the general claim.

The two obligations

Let P(n)P(n) be the statement to prove for every integer n≥n0n\ge n_0. An induction proof establishes:

  1. Base case: P(n0)P(n_0) is true.
  2. Inductive step: for an arbitrary integer n≥n0n\ge n_0, assume P(n)P(n) and deduce P(n+1)P(n+1).
These establish P(n)P(n) for all integers n≥n0n\ge n_0. The starting value might be zero, one or another integer; choose it to match the claim. The logical quantifier “for every” in the second obligation matters: proving only the transition from four to five leaves the other transitions unproved. The University of Toronto’s induction guide sets out this base-case/step structure.

Assuming P(n)P(n) inside the step is legitimate because the immediate task is to prove a conditional implication. Assuming P(n+1)P(n+1), the conclusion of that implication, would leave the task undone.

A discounted sum, proved once

Consider a hypothetical stream paying one monetary unit at dates 0,1,…,n−10,1,\ldots,n-1, with a stipulated discount factor of 1/21/2 per period. The date-zero payment is undiscounted. This deliberately simple factor is an arithmetic example, not an observed market rate.

Its present value is a finite geometric sum. Define the claim
P(n):Sn=∑k=0n−1(12)k=2(1−2−n),n≥1.P(n):\qquad S_n=\sum_{k=0}^{n-1}\left(\frac12\right)^k =2\left(1-2^{-n}\right),\qquad n\ge1.
Base case, n=1n=1. The single payment is worth S1=1S_1=1, and the proposed expression gives 2(1−1/2)=12(1-1/2)=1.
Inductive step. Assume Sn=2(1−2−n)S_n=2(1-2^{-n}) for an arbitrary integer n≥1n\ge1. Adding the next payment gives
Sn+1=Sn+2−n=2(1−2−n)+2−n=2−2−n=2(1−2−(n+1)).\begin{aligned} S_{n+1} &=S_n+2^{-n}\\ &=2(1-2^{-n})+2^{-n}\\ &=2-2^{-n}\\ &=2\left(1-2^{-(n+1)}\right). \end{aligned}

The second line is exactly where the induction hypothesis enters. The last line is the required formula for n+1n+1. Together with the base case, this proves the identity for every positive integer payment count. □\square

For four payments, 1+0.5+0.25+0.125=1.8751+0.5+0.25+0.125=1.875 monetary units. A fifth adds 0.06250.0625, giving 1.93751.9375. Those are useful arithmetic checks, while the arbitrary-nn argument supplies the proof.

A base case P(1) and a general implication from P(n) to P(n+1) establish every positive integer case. Five finite geometric sums are 1, 1.5, 1.75, 1.875 and 1.9375 present-value units; the five checks alone do not prove the identity.
A base case P(1) and a general implication from P(n) to P(n+1) establish every positive integer case. Five finite geometric sums are 1, 1.5, 1.75, 1.875 and 1.9375 present-value units; the five checks alone do not prove the identity.
Original exact arithmetic for the hypothetical discount factor 1/2. The chain describes the proof obligations; the bars illustrate only five finite cases.
The update Sn+1=Sn+2−nS_{n+1}=S_n+2^{-n} is a recurrence relation. Induction verifies that the proposed closed form satisfies both that update and its starting value. This is useful when replacing a loop over payments with a direct expression: the proof connects the two mathematical calculations, although software still needs correct indexing and numerical arithmetic.

What the proof does not give you

A step without a base proves nothing about whether the chain starts. The false identity
1+2+⋯+n=n(n+1)2+11+2+\cdots+n=\frac{n(n+1)}{2}+1

survives the formal “add n+1n+1 to both sides” manipulation. Yet at n=1n=1 its two sides are 11 and 22. The failed base case prevents an induction proof.

All finite counts do not include an infinite count. For a general real ratio q≠1q\ne1, the finite identity is ∑k=0n−1qk=(1−qn)/(1−q)\sum_{k=0}^{n-1}q^k=(1-q^n)/(1-q). Replacing it with an infinite sum requires a separate limit argument. For ∣q∣<1|q|<1, qn→0q^n\to0 and the sum tends to 1/(1−q)1/(1-q); for q=2q=2, the finite identity remains valid but the infinite series diverges. See the University of Waterloo’s finite and infinite geometric-series discussion.
Some steps need more than the immediately preceding case. Strong induction assumes all cases from the starting index through nn when proving n+1n+1. It is convenient for recurrences involving several earlier terms. It has the same logical power as ordinary induction: apply ordinary induction to the accumulated statement that all cases up to nn hold. MIT’s induction chapter, especially §5.3 explains this equivalence.
For the separate question of when partial sums converge, SelfMadeQuant’s Series and the Geometric Series: Partial Sums and Convergence develops the limit argument and financial applications. The public page is a preview; the full lesson requires membership.

Sources

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  • mathematical induction
  • proof
  • geometric series
  • recurrence relations
  • quantitative foundations
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