Geometric and telescoping series · Calculus II: Integrals & Series

§7.2 defined series convergence but only found a closed form for the partial sums in one lucky case, \sum\frac1{2^n}. This lesson generalizes that case into a full family — geometric series, where a clean formula always exists — and introduces a second family, telescoping series, where the closed form comes from massive cancellation instead of a formula.

Geometric series

A geometric series has the form

$\sum_{n=0}^\infty ar^n=a+ar+ar^2+ar^3+\cdots$

— each term is the previous one times a fixed common ratio r. Its partial sum has a classic closed form, derived with a trick: write S_N, multiply by r, and subtract to cancel every middle term:

$S_N=a+ar+\cdots+ar^{N-1}$ $rS_N=\quad\ ar+ar^2+\cdots+ar^{N-1}+ar^N$ $S_N-rS_N=a-ar^N\ \Longrightarrow\ S_N=\frac{a(1-r^N)}{1-r}\quad(r\ne1)$

Convergence depends entirely on r^N. From §7.0's exponential-growth reasoning:

$r^N\to\begin{cases}0&|r|<1\\ \text{diverges}&|r|\ge1,\ r\ne1\\ =1\text{ always}&r=1\end{cases}$

So for |r|<1:

$\sum_{n=0}^\infty ar^n=\lim_{N\to\infty}\frac{a(1-r^N)}{1-r}=\frac{a}{1-r}$

For |r|\ge1, the series diverges (for r=1, S_N=aN\to\pm\infty; for r=-1, S_N oscillates like §7.2's worked example; for |r|>1, |ar^N| itself grows without bound, so the terms don't even shrink to zero — failing §7.2's necessary condition outright).

§7.2's \sum\frac1{2^n} is this formula with a=\frac12, r=\frac12 (summing from n=1 instead of n=0 just shifts the index): \frac{1/2}{1-1/2}=1 — matching exactly.

Telescoping series

A telescoping series is built so that consecutive terms cancel when written as a sum of partial sums — usually because a partial-fraction decomposition (§4.9) splits each term into a difference:

$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$

Write out the partial sum and watch the middle terms cancel:

$S_N=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)$

Every term except the very first (1) and the very last (-\frac1{N+1}) cancels against a neighbor:

$S_N=1-\frac1{N+1}$

$\sum_{n=1}^\infty\frac1{n(n+1)}=\lim_{N\to\infty}\left(1-\frac1{N+1}\right)=\boxed1$

The technique generalizes: any series whose terms decompose as b_n-b_{n+1} for some sequence b_n telescopes to S_N=b_1-b_{N+1}, and the whole series converges exactly when b_n itself converges — reducing an infinite-sum question back to a single sequence limit, §7.2's founding idea in its purest form.

Doing it in Python

The geometric series formula, confirmed against a direct partial-sum computation:

from fractions import Fraction

a, r = Fraction(3), Fraction(1, 4)

formula = a / (1 - r)
print(f"formula: a/(1-r) = {formula} = {float(formula)}")

S_N = sum(a * r**n for n in range(200))   # 200 terms, r < 1 so this is already extremely close
print(f"200-term partial sum        = {float(S_N):.10f}")

The telescoping series \sum\frac1{n(n+1)}, shown decomposing and cancelling explicitly for a handful of terms:

from fractions import Fraction

N = 6
terms = [(Fraction(1, n), Fraction(1, n + 1)) for n in range(1, N + 1)]
for n, (lo, hi) in enumerate(terms, start=1):
    print(f"n={n}: 1/{n} - 1/{n+1} = {lo} - {hi}")

S_N = sum(lo - hi for lo, hi in terms)
print(f"\nS_{N} = {S_N}  (matches 1 - 1/(N+1) = {1 - Fraction(1, N+1)})")

A geometric series that diverges (r>1), confirmed by watching the partial sums grow without bound rather than settle:

r = 1.5
print(f"{'N':>4} {'S_N':>12}")
for N in (5, 10, 20, 40):
    S_N = sum(r**n for n in range(N))
    print(f"{N:>4} {S_N:>12.2f}")
print("\nno sign of settling down -- diverges, since |r| > 1")

Worked example

Find the sum \displaystyle\sum_{n=0}^\infty3\left(\frac14\right)^n.

This is geometric with a=3, r=\frac14. Since |r|=\frac14<1, it converges:

$\sum_{n=0}^\infty3\left(\frac14\right)^n=\frac{a}{1-r}=\frac3{1-\frac14}=\frac3{3/4}=\boxed4$

Sanity check. The first term alone is 3, and every subsequent term is positive, so the sum must exceed 3 — and 4 does. The terms shrink by a factor of 4 each step (3,0.75,0.1875,\ldots), so almost the entire sum comes from the first couple of terms: 3+0.75=3.75, already 93.75\% of the way to 4 after just two terms. ✓

Your turn

1. Find \displaystyle\sum_{n=0}^\infty\left(-\frac13\right)^n.

2. A ball is dropped from height 10 m and, on each bounce, rebounds to \frac35 of the height it fell from. Find the total vertical distance it travels (down and up, summed over infinitely many bounces). Hint: the first drop is 10 m down; every bounce after that contributes a rise and a fall of the same, shrinking height — a geometric series.

3. Find \displaystyle\sum_{n=1}^\infty\left(\frac1{n+1}-\frac1{n+2}\right) by recognizing the telescoping structure directly, without expanding terms.

Solutions

1. a=1, r=-\frac13, and |r|=\frac13<1:

$\sum_{n=0}^\infty\left(-\frac13\right)^n=\frac1{1-(-1/3)}=\frac1{4/3}=\boxed{\frac34}$

2. The first fall contributes 10 m. Every bounce after that rises 10\left(\frac35\right) and falls the same distance, so contributes 2\cdot10\left(\frac35\right)^n for n=1,2,3,\ldots Total distance:

$D=10+\sum_{n=1}^\infty2\cdot10\left(\frac35\right)^n=10+20\cdot\frac{3/5}{1-3/5}=10+20\cdot\frac{3/5}{2/5}=10+20\cdot\frac32$

$=10+30=\boxed{40\text{ m}}$

A ball dropped from 10 m, bouncing forever with a \frac35 rebound ratio, travels a finite total distance — a physical echo of §7.2's "infinitely many positive terms, finite sum" surprise.

3. This is \sum(b_n-b_{n+1}) with b_n=\frac1{n+1}, so it telescopes to b_1-\lim_{n\to\infty}b_{n+1}=\frac12-0=\boxed{\frac12} — no term-by-term expansion needed, just matching the pattern from the concept section.

Check yourself in code

Compute the geometric series \sum_{n=0}^\infty3\left(\frac14\right)^n using the formula \frac{a}{1-r}, and the telescoping series \sum_{n=1}^\infty\frac1{n(n+1)} using SymPy's summation.

Print exactly this:

geometric sum   = 4
telescoping sum = 1

import sympy as sp

n = sp.Symbol('n')

a, r = sp.Integer(3), sp.Rational(1, 4)
geometric = a / (1 - r)
print("geometric sum   = ...")

telescoping = sp.summation(1 / (n * (n + 1)), (n, 1, sp.oo))
print("telescoping sum = ...")

import sympy as sp

n = sp.Symbol('n')

a, r = sp.Integer(3), sp.Rational(1, 4)
geometric = a / (1 - r)
print(f"geometric sum   = {geometric}")

telescoping = sp.summation(1 / (n * (n + 1)), (n, 1, sp.oo))
print(f"telescoping sum = {telescoping}")

Geometric series always have a closed-form partial sum, S_N=\frac{a(1-r^N)}{1-r}, converging to \frac{a}{1-r} exactly when |r|<1 — the single most important convergence test threshold in this module, and one that will reappear as the conclusion of both the ratio test (§7.9) and the discussion of power series (Module 8). Telescoping series get their closed form differently, from a partial-fraction decomposition that cancels almost everything, collapsing an infinite sum back to a single sequence limit.

Next: the geometric series' most famous application — resolving Zeno's 2,500-year-old paradoxes about motion, once and for all.

29. Geometric and telescoping series