Phase 2 - Lesson 2.7

Sequences, Series, Convergence, and Power Series

When an infinite sum has a finite value, the tests that decide it, and power series - the bridge from geometric discounting to Taylor expansions.

⏱ 60 min● Intermediate🔗 Prereqs: 2.6

↖ Phase 2 hub

Builds on: 1.5 defined sequence limits; a series is the limit of the sequence of its partial sums.
Leads to: Power series formalize Taylor expansions (2.4); geometric series price perpetuities and annuities and recur throughout fixed income.

Learning Objectives

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Define convergence of a series via its partial sums and evaluate a geometric series.
Apply the n-th term, ratio, and comparison tests to decide convergence.
Distinguish absolute from conditional convergence.
Find the radius of convergence of a power series and connect geometric series to present value.

Key Vocabulary

Series: The formal infinite sum ∑ a_n; it converges to S if its partial sums s_N=∑_{n≤N} a_n → S.

Geometric series: ∑ ar^n; converges iff |r|<1, to a/(1−r).

n-th term test: If a_n does not → 0, the series diverges (a necessary, not sufficient, condition).

Ratio test: If lim|a_{n+1}/a_n|=L, the series converges (absolutely) for L<1, diverges for L>1.

Absolute convergence: ∑|a_n| converges; it implies convergence and permits rearrangement.

Radius of convergence: The R such that a power series ∑c_n(x−a)^n converges for |x−a|<R.

A series is the limit of its partial sums

The infinite sum $\sum_{n=0}^{\infty} a_n$ means the limit of the partial sums $s_N=\sum_{n=0}^{N}a_n$. If that limit exists and is finite, the series converges; otherwise it diverges. The cleanest example is geometric:

\[\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}\quad\text{provided}\ |r|\lt 1\] (2.10)

Worked Example - Why the geometric formula holds

Partial sum: $s_N=a+ar+\cdots+ar^N$. Multiply by $r$ and subtract: $s_N(1-r)=a(1-r^{N+1})$.

So $s_N=\dfrac{a(1-r^{N+1})}{1-r}$.

If $|r|\lt 1$ then $r^{N+1}\to0$, giving $s_N\to\dfrac{a}{1-r}$. If $|r|\ge1$ the terms do not shrink and it diverges.

Tests for convergence

The n-th term test is the first filter: if $a_n\not\to0$, the series diverges. But $a_n\to0$ is not enough - the harmonic series $\sum 1/n$ has terms going to 0 yet diverges. Two workhorses decide the rest:

Ratio test: compute $L=\lim|a_{n+1}/a_n|$; converges if $L\lt 1$, diverges if $L\gt 1$, inconclusive if $L=1$.
Comparison test: if $0\le a_n\le b_n$ and $\sum b_n$ converges, so does $\sum a_n$ (and the contrapositive for divergence).

A series is absolutely convergent if $\sum|a_n|$ converges; absolute convergence implies convergence and lets you rearrange terms freely. The alternating harmonic series $\sum(-1)^{n+1}/n$ converges but only conditionally - a distinction that matters for rearrangement.

Power series and Taylor

A power series $\sum_{n=0}^{\infty} c_n(x-a)^n$ converges on an interval $|x-a|\lt R$, where the radius of convergence $R$ often comes from the ratio test. Taylor series (2.4) are power series whose coefficients are $c_n=f^{(n)}(a)/n!$; where they converge to $f$, the function is analytic.

Finance link: an annuity as a geometric series

A stream of equal payments $C$ at the end of each year for $n$ years, discounted at per-period rate $i$, is a finite geometric series with ratio $v=1/(1+i)$:

\[\mathrm{PV}=\sum_{k=1}^{n} \frac{C}{(1+i)^k}=C\cdot\frac{1-(1+i)^{-n}}{i}\] (2.11)

Letting $n\to\infty$ gives the perpetuity $\mathrm{PV}=C/i$: the infinite geometric series converges precisely because the discount ratio satisfies $|v|\lt 1$. Discounting is a geometric series.

Interactive: partial sums approaching the limit

Common Mistakes to Avoid

Concluding convergence from a_n→0; that is necessary only. The harmonic series is the standing counterexample.
Applying the geometric formula a/(1−r) when |r|≥1, where the series diverges.
Using the ratio test’s L=1 case as a verdict - it is inconclusive; switch tests.
Rearranging a conditionally convergent series and expecting the same sum (it can be changed to anything).

Quant Practitioner Tips

Run the n-th term test first - it is free and instantly kills many divergent series.
The ratio test is the default for factorials and exponentials in the terms.
Absolute convergence is the ‘safe’ kind: it survives rearrangement and underlies interchange of sum and integral.
Recognize discounting as a geometric series; perpetuity C/i and annuity formulas fall straight out.

Practice this in the Euler Lab

Computational problems that exercise exactly this technique. Each opens in the Euler Lab with a Python workbench, a progressive hint ladder, and answer checking. Tier A/B run at full scale in the browser.

Warm-up:
#2 Even Fibonacci Numbers (1%, tier A) #6 Sum Square Difference (1%, tier A) #42 Coded Triangle Numbers (2%, tier A) #12 Highly Divisible Triangular Number (3%, tier A)

Applied:
#313 Sliding Game (16%, tier B) #323 Bitwise-OR Operations on Random In (18%, tier B) #713 Turán's Water Heating System (19%, tier B)

Challenge:
#208 Robot Walks (41%, tier C) #375 Minimum of Subsequences (41%, tier C)

183 Project Euler problems in total are mapped to this lesson. Open the Euler Lab to filter them all.

Knowledge Check

Q1 Easy

The geometric series ∑ ar^n converges if and only if:

a≠0

|r|<1

r>0

a<1

Q2 Medium

Which statement is correct about the harmonic series ∑1/n?

it converges because 1/n→0

it diverges even though 1/n→0

the ratio test proves it converges

it converges to 1

Q3 Hard

For the power series ∑ x^n/n!, the ratio test gives L=lim|x/(n+1)|=0, so it converges:

only for |x|<1

for all x (radius R=∞)

only at x=0

for x>0

Practical Exercise

(a) Determine whether $\sum_{n=1}^{\infty}\frac{2^n}{n!}$ converges, using the ratio test. (b) An annuity pays $1{,}000 at the end of each year for 5 years at $i=8\%$. Compute its present value as a geometric sum.

▶ Show full solution

(a) $\dfrac{a_{n+1}}{a_n}=\dfrac{2^{n+1}/(n+1)!}{2^n/n!}=\dfrac{2}{n+1}\to0$. Since $L=0\lt 1$, the series converges absolutely (its sum is in fact $e^2-1$).

(b) With $v=1/1.08$:

\[\mathrm{PV}=1000\cdot\frac{1-(1.08)^{-5}}{0.08}=1000\cdot\frac{1-0.68058}{0.08}\approx 1000\times 3.9927=\$3{,}992.7.\]

So the five payments are worth about $\$3{,}993$ today - a finite geometric series summed with the annuity formula.

After the reveal, answer for yourself: The same ratio-test machinery that classifies ∑2^n/n! also underlies why the annuity’s geometric sum is finite. Convergence theory and discounting are one subject.

Lesson Summary

A series converges when its partial sums do; the geometric series ∑ar^n sums to a/(1−r) for |r|<1. The n-th term, ratio, and comparison tests decide convergence, with absolute convergence the robust kind. Power series converge on an interval of radius R and formalize Taylor expansions. In finance, discounting is a geometric series - annuities and the perpetuity C/i drop straight out.

Formula Sheet Additions

Geometric series

\[\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r},\quad |r|\lt 1\]

The convergent infinite sum behind perpetuities and discounting.

Annuity present value

\[\mathrm{PV}=C\cdot\frac{1-(1+i)^{-n}}{i}\]

A finite geometric sum of discounted level payments.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: Why is a_n→0 necessary but not sufficient for ∑a_n to converge? Give the counterexample.
A: If the terms did not vanish the partial sums could not settle, so a_n→0 is required; but the harmonic series ∑1/n has terms →0 and still diverges, showing it is not sufficient.

Q: State the geometric-series value and connect it to a perpetuity.
A: ∑_{n≥0} ar^n=a/(1−r) for |r|<1; with discount ratio v=1/(1+i) a perpetuity paying C forever is C·v/(1−v)=C/i, a convergent geometric series.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

Source References

This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.

Calculus, Vol. I (Tom Apostol, 2nd ed., 1967) foundational - Vol. I, Ch. 10 - Infinite series, convergence tests, and power series with radius of convergence.
Probability: Theory and Examples (Rick Durrett, 5th ed.) current - Ch. 1 appendix - Series and absolute convergence as the analytic backbone for interchanging sums, integrals, and expectations.

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