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Mathematical Foundations for Quant Finance
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    1. PG5.1 A series is the limit of its partial sumsProof · 40 min
    2. PG5.2 The comparison and ratio tests, provedProof · 45 min
    3. PG5.3 Absolute and conditional convergenceProof · 40 min
    4. PG5.4 The series for eˣ converges for every xProof · 45 min
    5. PG5.5 Stage review: proving a series convergesProof · 40 min
Proof Gym overview →

Proof Gym PG5 Series and convergence Lesson 1 of 5

ProofLesson PG5.1 in the Proof Gym, stage PG5: Series and convergence. About 40 minutes.

A series is the limit of its partial sums

Builds onPG4 ε–δ limits1.5.3 Infinite series and when they converge

Simulate

Check the claim with numbers

Lesson 1.5.3 defined the sum of a series as the limit of its partial sums Sn=a1+a2+⋯+anS_n = a_1 + a_2 + \cdots + a_n. Each term is the step from one partial sum to the next: an=Sn−Sn−1a_n = S_n - S_{n-1}. If the partial sums settle, their steps should shrink.

Test that on four series. For each, the cell prints S1000S_{1000}, S1001S_{1001} and the step between them, the term a1001a_{1001}.

⌘/Ctrl + Enter runs the cell

In the first three series, S1000S_{1000} and S1001S_{1001} agree to five decimal places or more, because the step a1001a_{1001} is tiny. The last series, 1−1+1−⋯1 - 1 + 1 - \cdots, jumps from 0 to 1. Its terms never shrink, and its partial sums never settle.

Conjecture

If the series ∑k=1∞ak\sum_{k=1}^{\infty} a_k converges, then its terms tend to 0:

an→0as n→∞a_n \to 0 \quad \text{as } n \to \infty

Four series agree. That is evidence, not a proof: the claim is about every series, and about a limit, which no finite list of terms can reach.

What "a series converges" means

A series converges when the sequence of its partial sums converges. Lesson PG4.1 made that precise with a tolerance ε\varepsilon and a place NN, so the same words now apply to SnS_n.

Rule: Convergence of a series

∑k=1∞ak=S means:for every ε>0 there isa whole number N with∣Sn−S∣<ε for every n>N\begin{gathered} \sum_{k=1}^{\infty} a_k = S \text{ means:} \\ \text{for every } \varepsilon > 0 \text{ there is} \\ \text{a whole number } N \text{ with} \\ |S_n - S| < \varepsilon \text{ for every } n > N \end{gathered}

SS is the sum of the series. A proof of convergence names NN for each ε\varepsilon, or shows that one exists. If every n>Nn > N works, so does every n≥N+1n \ge N + 1, so later lessons also write "for every n≥Nn \ge N".

Try it

Take 12+14+18+⋯\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots. Its partial sums are Sn=1−12nS_n = 1 - \frac{1}{2^n} (lesson 1.5.3), so ∣Sn−1∣=12n|S_n - 1| = \frac{1}{2^n}. That is below ε\varepsilon once 2n>1ε2^n > \frac{1}{\varepsilon}.

For ε=0.001\varepsilon = 0.001, the first such nn is 10, so N=9N = 9 works. The cell checks every nn from 10 to 60, and shows the miss at n=9n = 9 is too large.

⌘/Ctrl + Enter runs the cell

The miss is 0.001953125 after 9 terms and 0.0009765625 after 10. The proof below also needs the triangle inequality of lesson PG4.2: ∣u+v∣≤∣u∣+∣v∣|u + v| \le |u| + |v| for any numbers uu and vv.

Format 1

Reorder the steps

Here is the proof of the claim, with its lines shuffled. Put them in order with the arrows, or drag a line by its handle.

  1. By the definition, there is a whole number NN with ∣Sm−S∣<ε2|S_m - S| < \frac{\varepsilon}{2} for every m>Nm > N.
  2. Take any n>N+1n > N + 1. Then n>Nn > N and n−1>Nn - 1 > N, so SnS_n and Sn−1S_{n-1} are both within ε2\frac{\varepsilon}{2} of SS.
  3. Suppose ∑ak\sum a_k converges to SS, and let ε>0\varepsilon > 0.
  4. So ∣an∣<ε2+ε2=ε|a_n| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon for every n>N+1n > N + 1.
  5. For this nn, an=Sn−Sn−1=(Sn−S)+(S−Sn−1)a_n = S_n - S_{n-1} = (S_n - S) + (S - S_{n-1}), so ∣an∣≤∣Sn−S∣+∣S−Sn−1∣|a_n| \le |S_n - S| + |S - S_{n-1}|.
  6. This works for every ε>0\varepsilon > 0, with N+1N + 1 in place of NN, so an→0a_n \to 0.

Turned round, the claim becomes a test: if the terms do not tend to 0, the series diverges. This is the term test. It can prove divergence, but never convergence, as the next series shows.

Simulate

Shrinking terms are not enough

The harmonic series 1+12+13+⋯1 + \frac{1}{2} + \frac{1}{3} + \cdots has terms that tend to 0, so the term test says nothing. Group its terms in blocks that end at powers of 2. Block 1 is 12\frac{1}{2}, block 2 is 13+14\frac{1}{3} + \frac{1}{4}, and block 3 runs from 15\frac{1}{5} to 18\frac{1}{8}.

⌘/Ctrl + Enter runs the cell

Each row prints jj, the number of terms in block jj, and the block's total. Every block adds at least 0.5: from 0.5 for block 1 up to 0.6893 for block 7. So each doubling of nn should add at least 12\frac{1}{2}.

⌘/Ctrl + Enter runs the cell

The columns are mm, 2m2^m, the partial sum S2mS_{2^m} and 1+m21 + \frac{m}{2}. The partial sum never falls below 1+m21 + \frac{m}{2}: after 220=10485762^{20} = 1048576 terms it is 14.4402, against 11.0.

Conjecture

For every whole number m≥0m \ge 0, the partial sums of the harmonic series satisfy

S2m≥1+m2S_{2^m} \ge 1 + \frac{m}{2}

So the harmonic series diverges, although its terms tend to 0.

Eight rows agree, but only up to a million terms. A bound that grows with mm for every mm would show the partial sums pass every number.

Format 2

Fill the gap

The proof, with its key steps left out. Type each missing piece as you would a quick answer, such as 2^j or m/3, then check.

The bound. Block jj runs from the term 12j−1+1\dfrac{1}{2^{j-1} + 1} to the term 12j\dfrac{1}{2^j}. It has 2j−2j−1=2^j - 2^{j-1} = terms.

The smallest term in block jj is its last, 12j\dfrac{1}{2^j}. So the block adds up to at least 2j−1×12j2^{j-1} \times \dfrac{1}{2^j}, which is .

S2mS_{2^m} is the first term, 1, followed by blocks 1 to mm. Adding the bounds gives S2m≥1+S_{2^m} \ge 1 + .

No limit. Suppose the partial sums had a limit SS. Take ε=1\varepsilon = 1 in the definition: there is a whole number NN with Sn<S+1S_n < S + 1 for every n>Nn > N.

Choose a whole number m≥Nm \ge N that is also at least . Since 2m>m≥N2^m > m \ge N, the bound gives S2m≥1+m2≥S+1S_{2^m} \ge 1 + \frac{m}{2} \ge S + 1.

That contradicts Sn<S+1S_n < S + 1 for n=2mn = 2^m. So the partial sums have no limit, and the harmonic series diverges. For m=20m = 20 the bound promises S220≥S_{2^{20}} \ge , below the 14.4402 that the simulation found.

How to type answers
  • Powers with ^, brackets round a power that is a sum: r^k, r^(k+1).
  • Fractions with / and brackets: (1-r)/(1+r). Multiply with * or side by side: 2k+1.
  • Press Enter in any box to check them all.

Format 3

Find the error

A proof that 1−1+1−1+⋯=121 - 1 + 1 - 1 + \cdots = \frac{1}{2}, the series whose partial sums jump between 1 and 0. One line claims something that is not true. Select it.

Lesson PG5.2 turns the bound in the harmonic proof into a method, the comparison test. It also uses the geometric series as a yardstick for other series.

Summary card

  • A series converges to SS when its partial sums do: for every ε>0\varepsilon > 0 there is NN with ∣Sn−S∣<ε|S_n - S| < \varepsilon for every n>Nn > N.
  • The term test: if ∑ak\sum a_k converges, then an→0a_n \to 0. So terms that do not tend to 0 mean divergence.
  • Terms that tend to 0 are not enough. The harmonic series has S2m≥1+m2S_{2^m} \ge 1 + \frac{m}{2}, so it diverges.
  • 1−1+1−⋯1 - 1 + 1 - \cdots has no sum. Algebra on a sum that does not exist proves nothing.
Next lessonPG5.2 The comparison and ratio tests, proved →

Free further reading

Optional. This lesson covers what you need; these add more examples and practice. All are free to read.