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

Click a status chip to cycle: Not started → In progress → Studied → Practiced → Needs review → Mastered.

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

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