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Mathematical Foundations for Quant Finance
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    1. PG4.1 What a limit means: ε and NProof · 40 min
    2. PG4.2 Limits of sums and productsProof · 40 min
    3. PG4.3 Limits of functions: ε and δProof · 40 min
    4. PG4.4 Supremum, infimum and completenessProof · 40 min
    5. PG4.5 Monotone sequences, and why (1 + 1/n)ⁿ convergesProof · 40 min
    6. PG4.6 Stage review: limits from the definitionProof · 40 min
Proof Gym overview →

Proof Gym PG4 ε–δ limits Lesson 5 of 6

ProofLesson PG4.5 in the Proof Gym, stage PG4: ε–δ limits. About 40 minutes.

Monotone sequences, and why (1 + 1/n)ⁿ converges

Builds onPG4.4 Supremum, infimum and completenessPG4.2 Limits of sums and productsPG3.5 Stage review: two proofs of your own1.4.2 Exponential functions and the number e

Simulate

Check the claim with numbers

Lesson 1.4.2 defined ee as the number that (1+1n)n\left(1 + \frac{1}{n}\right)^n settles on. That it settles at all is a claim, and it needs a proof. Track an=(1+1n)na_n = \left(1 + \frac{1}{n}\right)^n beside bn=(1+1n)n+1b_n = \left(1 + \frac{1}{n}\right)^{n+1}, which has one extra factor.

⌘/Ctrl + Enter runs the cell

The ana_n climb from 2 towards 2.718, and the bnb_n fall from 4 towards the same number. The gap between them shrinks, to 0.002717 at n=1000n = 1000. Exact fractions test the pattern on the first 151 terms:

⌘/Ctrl + Enter runs the cell

All three print True: every ana_n is below the next, every bnb_n is above the next, and each ana_n is below its bnb_n. Pushing decimals too far, though, gives nonsense:

⌘/Ctrl + Enter runs the cell

At n=1015n = 10^{15} the computer stores 1+1n1 + \frac{1}{n} with a tiny rounding error, and the huge power magnifies it into 3.035035. At 101610^{16} it stores exactly 1, and prints 1.0. The numbers break here, not the claim: it is the rounding trap of lesson 2.1.1.

Conjecture

(1+1n)n converges as n→∞\left(1 + \frac{1}{n}\right)^n \text{ converges as } n \to \infty

151 exact terms rise, and each stays below the falling bnb_n, so below b1=4b_1 = 4. A proof must show this for every nn, and that a sequence like this must have a limit.

Monotone and bounded

In this lesson a sequence is increasing if an+1≥ana_{n+1} \ge a_n for every nn. Lesson 1.5.1 asked for >>, but the theorem below works either way. A sequence is bounded above if some number MM has an≤Ma_n \le M for every nn.

An increasing sequence can only climb without end, or level off. If it is bounded above, it cannot climb without end. So it should level off at the smallest number above all its terms, their supremum (lesson PG4.4).

Rule: Monotone convergence

an increasing and bounded above⇒an→sup⁡{an:n≥1}\begin{gathered} a_n \text{ increasing and bounded above} \\ \Rightarrow a_n \to \sup \{ a_n : n \ge 1 \} \end{gathered}

The same proof, turned upside down, shows that a decreasing sequence bounded below tends to the infimum of its terms.

The theorem needs completeness, which gives the supremum without naming it. With fractions alone it fails. The decimals 1, 1.4, 1.41, 1.414, … of 2\sqrt{2} increase and stay below 2, but no fraction is their limit.

Format 1

Reorder the steps

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

  1. Let ana_n be increasing and bounded above. By completeness, the set of its terms has a supremum ss.
  2. Take any n>Nn > N. The sequence is increasing, so an≥aN>s−εa_n \ge a_N > s - \varepsilon.
  3. For this nn, also an≤sa_n \le s, because ss is an upper bound of the terms.
  4. As ε>0\varepsilon > 0 was arbitrary, an→sa_n \to s.
  5. So s−ε<an≤ss - \varepsilon < a_n \le s, which gives ∣an−s∣<ε|a_n - s| < \varepsilon for every n>Nn > N.
  6. Let ε>0\varepsilon > 0. Since s−ε<ss - \varepsilon < s, it is not an upper bound, so some term aNa_N has aN>s−εa_N > s - \varepsilon.

The sequence increases

Now prove the two facts the simulation found. Both compare neighbouring terms by their ratio, and both use Bernoulli's inequality from lesson PG3.5: (1+x)m≥1+mx(1 + x)^m \ge 1 + mx for every x≥−1x \ge -1 and whole number m≥1m \ge 1.

Format 2

Fill the gap

The proof that an=(1+1n)na_n = \left(1 + \frac{1}{n}\right)^n increases, with its key steps left out. Type each missing piece in nn, as you would a quick answer, then check.

Write an=(n+1n)na_n = \left(\frac{n + 1}{n}\right)^n. Then

an+1an=(n+2n+1)n+1(nn+1)n\frac{a_{n+1}}{a_n} = \left(\frac{n + 2}{n + 1}\right)^{n+1} \left(\frac{n}{n + 1}\right)^{n}

Give the second factor one more power: (nn+1)n\left(\frac{n}{n + 1}\right)^n is (nn+1)n+1\left(\frac{n}{n + 1}\right)^{n+1} times . Then the two powers combine:

an+1an=n+1n(n(n+2)(n+1)2)n+1\frac{a_{n+1}}{a_n} = \frac{n + 1}{n} \left(\frac{n(n + 2)}{(n + 1)^2}\right)^{n+1}

The fraction inside is 1−1 - , because n(n+2)=(n+1)2−1n(n + 2) = (n + 1)^2 - 1.

Use Bernoulli's inequality with x=−1(n+1)2x = -\frac{1}{(n + 1)^2}, which is at least −1-1, and the power m=n+1m = n + 1:

(1−1(n+1)2)n+1≥1−n+1(n+1)2\begin{gathered} \left(1 - \frac{1}{(n + 1)^2}\right)^{n+1} \\ \ge 1 - \frac{n + 1}{(n + 1)^2} \end{gathered}

The right side simplifies to . So an+1an≥n+1n⋅nn+1=\frac{a_{n+1}}{a_n} \ge \frac{n + 1}{n} \cdot \frac{n}{n + 1} = , and since every ana_n is positive, an+1≥ana_{n+1} \ge a_n.

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.

The sequence stays below 4

Format 2

Fill the gap: a bound

The proof that every ana_n is below 4, with its key steps left out. Type each missing piece in nn, or as a number, then check.

Write bn=(1+1n)n+1b_n = \left(1 + \frac{1}{n}\right)^{n+1}. It is ana_n times 1+1n1 + \frac{1}{n}, so an<bna_n < b_n. The same steps as above show that bnb_n falls:

bnbn+1=n+1n+2((n+1)2n(n+2))n+1\frac{b_n}{b_{n+1}} = \frac{n + 1}{n + 2} \left(\frac{(n + 1)^2}{n(n + 2)}\right)^{n+1}

The fraction inside is 1+1 + .

Bernoulli's inequality, with x=1n(n+2)x = \frac{1}{n(n + 2)} and the power n+1n + 1, makes the bracket at least 1+n+1n(n+2)1 + \frac{n + 1}{n(n + 2)}. That is more than 1+1n+11 + \frac{1}{n + 1}, because (n+1)2>n(n+2)(n + 1)^2 > n(n + 2), and 1+1n+1=1 + \frac{1}{n + 1} = .

So bnbn+1>n+1n+2⋅n+2n+1=1\frac{b_n}{b_{n+1}} > \frac{n + 1}{n + 2} \cdot \frac{n + 2}{n + 1} = 1, and each bn+1b_{n+1} is below bnb_n. Hence an<bn≤b1=a_n < b_n \le b_1 = for every nn.

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.

The limit is e

So ana_n is increasing and bounded above by 4. By monotone convergence it converges, to the supremum of its terms. That limit is the number ee of lesson 1.4.2, and since a1=2a_1 = 2 and 4 is an upper bound, 2≤e≤42 \le e \le 4.

The two sequences also trap ee. By the product rule (lesson PG4.2), bn=an(1+1n)→e⋅1=eb_n = a_n\left(1 + \frac{1}{n}\right) \to e \cdot 1 = e.

A falling sequence never drops below its limit. If some bNb_N were below ee, every later bnb_n would be below ee by at least e−bNe - b_N, so it could not tend to ee.

So an≤e≤bna_n \le e \le b_n for every nn. At n=1000n = 1000 the simulation's table gives 2.716924≤e≤2.7196412.716924 \le e \le 2.719641.

Format 3

Find the error

A proof that (1+1n)n→1(1 + \frac{1}{n})^n \to 1, which the simulation contradicts. One line uses a rule beyond its reach. Select it.

Summary card

  • Monotone convergence: an increasing sequence that is bounded above converges, to the supremum of its terms. It needs completeness.
  • an=(1+1n)na_n = \left(1 + \frac{1}{n}\right)^n increases, by Bernoulli's inequality applied to an+1an\frac{a_{n+1}}{a_n}.
  • bn=(1+1n)n+1b_n = \left(1 + \frac{1}{n}\right)^{n+1} decreases, by Bernoulli again, and an<bn≤b1=4a_n < b_n \le b_1 = 4.
  • So ana_n converges, and its limit is ee, with an≤e≤bna_n \le e \le b_n for every nn.
  • The product rule needs a fixed number of factors: 1∞1^\infty is not 1.
Next lessonPG4.6 Stage review: limits from the definition →

Free further reading

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