Geometric Infinite Sum · Rafael Braga-Kribitz

Contents

01Deep Breakdown
02Flowchart
03Real-world use
04The formula, piece by piece
05The formula in words
06Critical Concepts
07Step-by-step calculation (tiny example)
08Code examples

S_\infty = \frac{a}{1-r} \quad (|r| < 1)

(1)

Deep Breakdown

What it tells me (Situation Awareness): This gives the exact sum of infinitely many terms of a geometric series, when the common ratio |r| < 1 so each term shrinks and the total stays finite. In DA/DS this is the limiting form of the geometric sum as n → ∞. It appears in: price of a perpetuity (infinite cashflow stream), total discounted revenue from a customer who churns with constant probability each period (geometric distribution CDF), steady-state queue length in M/M/1 queues, total probability mass of geometric distribution ( $\sum_{k=0}^{\infty} (1-p)^k p = 1$ ), and regularization paths where penalties sum to a finite value.

Flowchart

Real-world use

Where it is used

Any model with an infinite horizon of geometrically decaying terms: customer lifetime value under constant churn, discounted utility in reinforcement learning (return = sum of gamma^t rewards), perpetuity pricing, geometric series convergence tests, and infinite product moments of geometric distributions.

Problem it solves

Summing to infinity sounds intractable but when |r| < 1 the infinite sum is finite and has a simple closed form. Without it you'd have to truncate at some large n and accept approximation error. The formula is exact for the infinite case.

How it solves the problem

Derived from the finite sum: $S_n = \frac{a(1-r^n)}{1-r}$ . As n → ∞ and |r| < 1, the term r^n → 0. So $S_\infty = \lim_{n \to \infty} S_n = \frac{a(1-0)}{1-r} = \frac{a}{1-r}$ .

Example in a business

Customer LTV under constant churn: A SaaS customer pays €200/month. Monthly churn probability = 5%, so monthly retention r = 0.95. Expected LTV (present value ignoring discount rate) = 200 / (1 - 0.95) = 200 / 0.05 = €4,000. This is the exact expected total revenue from one customer, assuming geometric churn distribution. Including a monthly discount rate d: effective ratio = r/(1+d). LTV = a / (1 - r/(1+d)) = a(1+d)/(1+d-r).

The formula, piece by piece

Breaking $S_\infty = \frac{a}{1-r}$ token by token:

$S_\infty$ — Sum to infinity. The limit of partial sums as n → ∞.
$a$ — First term. The initial value of the sequence.
$1-r$ — Denominator: the "gap" between 1 and the ratio. As r → 1 from below, this gap → 0 and the sum → ∞. As r → 0, sum → a (only the first term matters).
Convergence condition $|r| < 1$ — Mandatory. If |r| ≥ 1 the terms don't shrink and the series diverges.
Special cases:
r = 0: S = a (single term, all others zero) - r = 0.5: S = 2a (sum equals twice the first term) - r = -0.5: S = a/1.5 = 2a/3 (alternating, converges to less than first term)

The formula in words

"The infinite sum of a converging geometric series equals the first term divided by (one minus the common ratio)." Intuition: the first term a sets the scale, and 1/(1-r) is the "leverage multiplier" from the infinite tail. When r = 0.9, the multiplier is 10; when r = 0.5, it's 2; when r = 0.1, it's 1.11.

Critical Concepts

Convergence requires |r| < 1 strictly: r = 1 gives a constant series that sums to infinity. r = -1 gives alternating ±a that never settles.
Negative r: valid — the series alternates and converges. e.g., r = -0.5: terms are a, -0.5a, 0.25a, -0.125a, ... sum = a/1.5 = 2a/3.
Geometric distribution connection: P(X = k) = (1-p)·p^k. Sum over k=0 to ∞: a=1-p, r=p → total = (1-p)/(1-p) = 1. Confirms proper probability distribution.
Infinite series vs. finite: Use finite formula $a(1-r^n)/(1-r)$ when you have n periods. Use infinite formula when modeling "forever" or when n is large enough that the truncation error is negligible (error ≈ a·r^n/(1-r)).
Discount-adjusted perpetuity: NPV of perpetuity with cashflow C and discount rate d per period = C/d.

This comes from r = 1/(1+d), giving S = C/(1+d) / (1 - 1/(1+d)) = C/d.

Step-by-step calculation (tiny example)

Series: a = 1, r = 0.5 (terms: 1, 0.5, 0.25, 0.125, ...)

Step 1: Verify convergence |r| = 0.5 < 1 → converges.

Step 2: Apply formula S_∞ = 1 / (1 - 0.5) = 1 / 0.5 = 2

Step 3: Verify with partial sums S_1 = 1 S_2 = 1.5 S_3 = 1.75 S_4 = 1.875 S_10 = 1 × (1 - 0.5^10)/(0.5) = (1 - 0.000977)/0.5 ≈ 1.998 The partial sums approach 2 from below.

Step 4: Interpret Every doubling of terms adds half the remaining gap to 2. This is the geometric convergence rate.

Code examples

Beginner example (Python)

python

"""
Geometric infinite sum: S_inf = a / (1 - r), |r| < 1

Used here to compute expected customer LTV under constant monthly churn.
"""
import numpy as np


def geometric_infinite_sum(first_term: float, ratio: float) -> float:
    """Return sum of infinite geometric series.

    Parameters
    ----------
    first_term : float
        First term a.
    ratio : float
        Common ratio r. Must satisfy |r| < 1.

    Returns
    -------
    float
        S_inf = first_term / (1 - ratio)

    Raises
    ------
    ValueError
        If |ratio| >= 1 (series diverges).
    """
    if abs(ratio) >= 1.0:
        raise ValueError(f"Series diverges: |r| = {abs(ratio):.4f} >= 1")
    return first_term / (1.0 - ratio)


# SaaS customer LTV: €200/month, 5% monthly churn → r = 0.95
MONTHLY_REVENUE = 200.0
MONTHLY_CHURN = 0.05
RETENTION_RATE = 1.0 - MONTHLY_CHURN
ltv = geometric_infinite_sum(MONTHLY_REVENUE, RETENTION_RATE)
print(f"Expected LTV (no discount): €{ltv:,.0f}")

# With monthly discount rate d = 1%
MONTHLY_DISCOUNT = 0.01
effective_ratio = RETENTION_RATE / (1.0 + MONTHLY_DISCOUNT)
ltv_discounted = geometric_infinite_sum(MONTHLY_REVENUE / (1.0 + MONTHLY_DISCOUNT), effective_ratio)
print(f"Expected LTV (discounted) : €{ltv_discounted:,.0f}")

# Verify with finite sum at n=120 months
partial_sum = sum(MONTHLY_REVENUE * RETENTION_RATE**k for k in range(120))
print(f"120-month partial sum : €{partial_sum:,.0f} (≈ LTV, small tail remaining)")

Beginner example (R)

# Geometric infinite sum: S_inf = a / (1 - r), for |r| < 1
# Application: discount factor sum in reinforcement learning
geometric_infinite_sum <- function(first_term, ratio) {
  # Infinite geometric series sum
  # Args:
  #   first_term : first value in series (a)
  #   ratio      : common ratio (r), must have |r| < 1
  # Returns: exact infinite sum
  if (abs(ratio) >= 1) {
    stop(sprintf("Series diverges: |r| = %.4f >= 1", abs(ratio)))
  }
  first_term / (1 - ratio)
}

# Reinforcement learning: total discounted return
# gamma = 0.99, constant reward r_t = 1 at every step
DISCOUNT_FACTOR <- 0.99
CONSTANT_REWARD <- 1.0
total_return <- geometric_infinite_sum(CONSTANT_REWARD, DISCOUNT_FACTOR)
cat(sprintf("Max possible discounted return (gamma=0.99): %.2f\n", total_return))

# Verify: partial sum at 1000 steps
steps <- 0:999
partial <- sum(CONSTANT_REWARD * DISCOUNT_FACTOR^steps)
cat(sprintf("1000-step partial sum : %.2f\n", partial))
cat(sprintf("Tail remaining (approx a*r^n/(1-r)) : %.4f\n", CONSTANT_REWARD * DISCOUNT_FACTOR^1000 / (1 - DISCOUNT_FACTOR)))

Professional example (Python)

python

"""
Infinite geometric series for perpetuity pricing and sensitivity analysis.

Business context: Fixed-income team pricing a perpetuity bond that pays coupon C
per period, with a discount rate d that varies by credit rating scenario.
Goal: compute fair value and duration (sensitivity to rate changes).
"""
import numpy as np
import pandas as pd

ANNUAL_COUPON_EUR: float = 1_000.0
FACE_VALUE_EUR: float = 10_000.0
CREDIT_SCENARIOS: dict[str, float] = {
    "AAA (2%)": 0.02,
    "AA (3%)": 0.03,
    "A (5%)": 0.05,
    "BBB (7%)": 0.07,
    "BB (10%)": 0.10,
}


def perpetuity_price(annual_coupon: float, discount_rate: float) -> float:
    """Price a perpetuity as geometric infinite sum.

    The perpetuity pays annual_coupon at end of each year forever.
    Price = sum_{t=1}^{inf} C / (1+d)^t
          = C/(1+d) * sum_{k=0}^{inf} (1/(1+d))^k   [geometric series]
          = C/(1+d) * 1/(1 - 1/(1+d))
          = C / d

    Parameters
    ----------
    annual_coupon : periodic payment
    discount_rate : per-period discount rate d (e.g. 0.05 for 5%)

    Returns
    -------
    float : present value of perpetuity
    """
    return annual_coupon / discount_rate


def modified_duration(discount_rate: float) -> float:
    """Duration of a perpetuity = (1 + d) / d."""
    return (1.0 + discount_rate) / discount_rate


rows = []
for scenario, rate in CREDIT_SCENARIOS.items():
    price = perpetuity_price(ANNUAL_COUPON_EUR, rate)
    duration = modified_duration(rate)
    yield_spread = rate - 0.02  # spread over AAA
    rows.append({
        "scenario": scenario,
        "discount_rate_pct": rate * 100,
        "price_eur": round(price, 2),
        "modified_duration_years": round(duration, 2),
        "yield_spread_bps": round(yield_spread * 10_000, 0),
    })

results_df = pd.DataFrame(rows)
print(results_df.to_string(index=False))

# Sensitivity: 1bp rate increase → price change ≈ -duration * price * 0.0001
print("\nPrice sensitivity to +1bp rate shock:")
for _, row in results_df.iterrows():
    price_change = -row["modified_duration_years"] * row["price_eur"] * 0.0001
    print(f" {row['scenario']}: €{price_change:+.2f} per bond")

Source: 1300 p. 313

#calculus #series-and-transforms #arithmetic-and-geometric-progressions