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Topos Uranos
English

The number e and the natural exponential

Summary

This lesson constructs the number ee as the supremum of the sequence (1+1n)n\left( 1 + \frac{1}{n} \right)^n, which arises from compound interest: it proves with the binomial theorem that the sequence increases and does not reach the value 33, and it encloses its supremum between this sequence and the decreasing sequence (1+1n)n+1\left( 1 + \frac{1}{n} \right)^{n + 1}. It then proves that ee is also the supremum of the sums of the reciprocals of the factorials, bounds the error of these sums, which makes it possible to compute ee to any desired precision, and proves that ee is irrational. Finally, it defines the natural exponential, relates it to the continuous compounding of interest and proves the inequality ex≥1+xe^x \geq 1 + x, which says that the line y=1+xy = 1 + x lies below the graph of the exponential.

Learning objectives

  1. Prove that the sequence (1+1n)n\left( 1 + \frac{1}{n} \right)^n is increasing and bounded, and define the number ee as its supremum.
  2. Compute approximations of ee by means of the sum of the reciprocals of the factorials, with a rigorous bound on the error, and prove that ee is irrational.
  3. Relate the natural exponential to compound interest and to its continuous compounding, and compare the effect of compounding at different frequencies.
  4. Prove and apply the inequality ex≥1+xe^x \geq 1 + x, and interpret it as the position of the line y=1+xy = 1 + x relative to the graph of y=exy = e^x.

The compound interest sequence

A capital of 11 is deposited for one year at an interest rate of one hundred percent per year. If the interest is paid only once, at the end of the year, the capital doubles and is worth 22. If it is paid twice, each half-year at half the rate, the interest of the first half-year itself earns interest in the second: the capital is multiplied twice by 1+121 + \frac{1}{2} and is worth 94\frac{9}{4}. With four quarterly payments it is worth (1+14)4=625256\left( 1 + \frac{1}{4} \right)^4 = \frac{625}{256}, a little more than 2.44, and with twelve monthly payments, a little more than 2.61. In general, if the year is divided into nn equal periods and in each one the fraction 1n\frac{1}{n} of the rate is paid, the capital is multiplied nn times by 1+1n1 + \frac{1}{n}, and at the end of the year it is worth

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

As nn increases, the capital is multiplied more times, but by a factor closer to 11, and it is not evident which of the two effects prevails: the powers of 11 are all equal to 11, while the powers of a fixed base greater than 11 grow without bound. This section proves that the sequence increases, but that its terms do not reach the value 33; its supremum is the number ee. The proofs rest on the binomial theorem from the lesson on combinatorics and the binomial theorem, on Bernoulli’s inequality, (1+x)n≥1+nx(1 + x)^n \geq 1 + nx for x≥−1x \geq -1, proved in the lesson on the Peano axioms, and on the completeness of the real numbers, stated in the lesson on number systems.

The growth of the sequence

By the binomial theorem, ana_n is a sum of n+1n + 1 terms. The term of index kk is (nk)1nk\binom{n}{k} \frac{1}{n^k}, and it is written as the product of 1k!\frac{1}{k!} and k−1k - 1 positive factors less than 11; with the convention that a product with no factors equals 11, the expression is also valid for k=0k = 0 and k=1k = 1:

an=∑k=0n1k!∏j=1k−1(1−jn)=1+1+12!(1−1n)+13!(1−1n)(1−2n)+⋯a_n = \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right) = 1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{n} \right) + \frac{1}{3!} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) + \cdots

In this expression, nn appears only in the factors 1−jn1 - \frac{j}{n}, which increase with nn, and in the number of terms, which also increases. The two effects push in the same direction; that is the idea of the theorem below.

TheoremGrowth of the sequence

For every natural number nn we have an<an+1a_n < a_{n + 1}, that is, the sequence (1+1n)n\left( 1 + \frac{1}{n} \right)^n is strictly increasing.

The binomial expansions of ana_n and of an+1a_{n + 1} are compared term by term. For n=3n = 3, the figure represents each term as a block in a column that starts at 22, because the first two terms are equal to 11 in every case.

Proof

  1. (nk)1nk=n(n−1)⋯(n−k+1)k! nk=1k!⋅nn⋅n−1n⋯n−k+1n\binom{n}{k} \frac{1}{n^k} = \frac{\resaltar{n(n - 1) \cdots (n - k + 1)}}{k! \, n^k} = \resaltar{\frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n - 1}{n} \cdots \frac{n - k + 1}{n}}

    By the binomial theorem, with x=1x = 1 and y=1ny = \frac{1}{n}, the term of index kk of ana_n is (nk)1nk\binom{n}{k} \frac{1}{n^k}. The binomial coefficient is a product of kk factors divided by k!k!, and each factor is divided by one of the kk factors nn of the denominator.

  2. an=∑k=0n1k!∏j=1k−1(1−jn)a_n = \dato{an}{\sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right)}

    The first quotient equals 11, and each of the ones after it is n−jn=1−jn\frac{n - j}{n} = 1 - \frac{j}{n}, for jj from 11 to k−1k - 1. Adding the terms gives the expression for ana_n.

  3. an+1=∑k=0n1k!∏j=1k−1(1−jn+1)+1(n+1)!∏j=1n(1−jn+1)a_{n + 1} = \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n + 1} \right) + \resaltar{\frac{1}{(n + 1)!} \prod_{j=1}^{n} \left( 1 - \frac{j}{n + 1} \right)}

    The same expression, with n+1n + 1 in place of nn, has one more term, the one of index k=n+1k = n + 1, which is positive because all its factors are.

  4. 0<1−jn<1−jn+1\dato{f}{0 < 1 - \resaltar{\frac{j}{n}} < 1 - \resaltar{\frac{j}{n + 1}}}

    For 1≤j≤n−11 \leq j \leq n - 1, as the denominator increases, jn\frac{j}{n} decreases, and the factor increases; moreover, all the factors are positive.

  5. 1k!∏j=1k−1(1−jn)<1k!∏j=1k−1(1−jn+1)\frac{1}{k!} \resaltar{\prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right)} < \frac{1}{k!} \resaltar{\prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n + 1} \right)}

    By the monotonicity of the product of positive numbers, each term of ana_n of index k≥2k \geq 2, if there is one, is less than the term of the same index of an+1a_{n + 1}; those of indices 00 and 11 are equal to 11 in both sums.

  6. a1=2<a2=94<a3=6427<a4=625256a_1 = 2 < a_2 = \frac{9}{4} < a_3 = \frac{64}{27} < a_4 = \frac{625}{256}

    Adding up, ana_n does not exceed the sum of the first n+1n + 1 terms of an+1a_{n + 1}, which still lacks the positive term of index n+1n + 1. Therefore, an<an+1a_n < a_{n + 1}. In the figure, the terms of the sequence rise.

An upper bound

Growth is not enough for the sequence to have a supremum: the natural numbers also grow. An upper bound is needed, and to obtain it ana_n is compared with the sum of the reciprocals of the factorials,

sn=∑k=0n1k!=1+1+12+16+⋯+1n!s_n = \sum_{k=0}^{n} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots + \frac{1}{n!}

which results from deleting, in the expression for ana_n, all the factors 1−jn1 - \frac{j}{n}. These sums are of interest in their own right, and the section that follows will prove that they approach the same number as ana_n, and much faster. Proposed problem 5 of the lesson on combinatorics asked for a proof that an<3a_n < 3; the second of the lemmas below contains this bound, together with the intermediate inequality an≤sna_n \leq s_n, which will be used later. The first compares the factorials with the powers of 22, and it is what makes it possible to bound sns_n by a geometric progression.

LemmaFactorials and powers of two

For every natural number kk we have k!≥2k−1k! \geq 2^{k - 1}, with equality only if k=1k = 1 or k=2k = 2.

The argument is by induction on kk, with the principle from the lesson on the Peano axioms: in passing from kk to k+1k + 1, the factorial is multiplied by k+1k + 1, which is at least 22, whereas the power only doubles. For kk from 11 to 44, the figure compares the factorial with the power, as two columns.

Proof

  1. 1!=1=201! = 1 = 2^{0}

    Base case: for k=1k = 1 both sides equal 11.

  2. k!≥2k−1\dato{h}{k! \geq 2^{k - 1}}

    Induction hypothesis: suppose that the inequality holds for a natural number kk.

  3. (k+1)!=(k+1)⋅k!≥2⋅2k−1=2k(k + 1)! = \resaltar{(k + 1)} \cdot k! \geq \resaltar{2} \cdot 2^{k - 1} = 2^{k}

    The factorial of k+1k + 1 is that of kk multiplied by k+1k + 1, which is at least 22; since all the factors are positive, the induction hypothesis gives the inequality for k+1k + 1.

  4. k!≥2k−1,3!=6>4=22k! \geq 2^{k - 1}, \qquad 3! = 6 > 4 = 2^2

    By the principle of induction, the inequality holds for every natural number kk. If k≥2k \geq 2, the factor k+1k + 1 is at least 33, and the inequality for k+1k + 1 is strict: equality occurs at k=1k = 1 and at k=2k = 2, and only there.

LemmaBound on the sequence

For every natural number nn we have an≤sn<3a_n \leq s_n < 3; the first inequality is strict for n≥2n \geq 2.

Each term of the expansion of ana_n is bounded by the corresponding term of sns_n, and each term of sns_n by one of a geometric progression with common ratio 12\frac{1}{2}. For n=4n = 4, the figure compares the three columns on the common base 22.

Proof

  1. 1k!∏j=1k−1(1−jn)≤1k!⋅1\frac{1}{k!} \resaltar{\prod_{j=1}^{k - 1} \left( 1 - \frac{j}{n} \right)} \leq \frac{1}{k!} \cdot \resaltar{1}

    Each factor 1−jn1 - \frac{j}{n} lies between 00 and 11; therefore their product does not exceed 11, and each term of ana_n does not exceed the term 1k!\frac{1}{k!} of sns_n.

  2. an≤sn\dato{as}{a_n \leq s_n}

    Adding for kk from 00 to nn gives the first inequality. If n≥2n \geq 2, the term of index 22 has the factor 1−1n<11 - \frac{1}{n} < 1, and the inequality is strict.

  3. 1k!≤12k−1,k≥1\dato{f}{\frac{1}{k!} \leq \frac{1}{2^{k - 1}}}, \quad k \geq 1

    By the lemma above, k!≥2k−1k! \geq 2^{k - 1} for every natural number kk; on taking reciprocals of two positive numbers, the inequality reverses.

  4. sn=1+∑k=1n1k!≤1+∑k=1n12k−1s_n = 1 + \resaltar{\sum_{k=1}^{n} \frac{1}{k!}} \leq 1 + \resaltar{\sum_{k=1}^{n} \frac{1}{2^{k - 1}}}

    We separate the term of index 00 and bound the others.

  5. 1+∑k=1n12k−1=1+2(1−12n)=3−12n−1<31 + \resaltar{\sum_{k=1}^{n} \frac{1}{2^{k - 1}}} = 1 + \resaltar{2 \left( 1 - \frac{1}{2^n} \right)} = 3 - \frac{1}{2^{n - 1}} < 3

    The highlighted expression is the sum of nn terms of a geometric progression with first term 11 and common ratio 12\frac{1}{2}, computed in the lesson on summations and progressions.

  6. an≤sn≤3−12n−1<3a_n \leq s_n \leq 3 - \frac{1}{2^{n - 1}} < 3

    Combining both inequalities gives the lemma. In the figure, with n=4n = 4, the geometric column stops 18\frac{1}{8} short of 33, and the other two lie below it.

A decreasing sequence that accompanies it

If the base 1+1n1 + \frac{1}{n} is raised to an exponent one greater, a second sequence is obtained, whose terms are greater than those of the first, because they are multiplied by a factor greater than 11:

bn=(1+1n)n+1=an(1+1n)>anb_n = \left( 1 + \frac{1}{n} \right)^{n + 1} = a_n \left( 1 + \frac{1}{n} \right) > a_n

Its first terms are b1=4b_1 = 4, b2=278b_2 = \frac{27}{8} and b3=25681b_3 = \frac{256}{81}: they go down. The theorem below proves this in general; the idea is to study the quotient of two consecutive terms, and the tool, Bernoulli’s inequality.

TheoremDecrease of the companion sequence

For every natural number nn we have bn>bn+1b_n > b_{n + 1}, that is, the sequence (1+1n)n+1\left( 1 + \frac{1}{n} \right)^{n + 1} is strictly decreasing.

It suffices to prove that the quotient bnbn+1\frac{b_n}{b_{n + 1}} is greater than 11, because the terms are positive. The figure shows the first terms of both sequences.

Proof

  1. bnbn+1=(n+1n)n+1(n+2n+1)n+2=((n+1)2n(n+2))n+1⋅n+1n+2\frac{b_n}{b_{n + 1}} = \frac{\left( \frac{n + 1}{n} \right)^{n + 1}}{\left( \frac{n + 2}{n + 1} \right)^{n + 2}} = \dato{c}{\left( \frac{(n + 1)^2}{n(n + 2)} \right)^{n + 1} \cdot \frac{n + 1}{n + 2}}

    We write each base as a quotient, split off a factor n+1n+2\frac{n + 1}{n + 2} from the power in the denominator and combine the two powers with exponent n+1n + 1 into a single one.

  2. (n+1)2n(n+2)=n(n+2)+1n(n+2)=1+1n(n+2)\frac{\resaltar{(n + 1)^2}}{n(n + 2)} = \frac{\resaltar{n(n + 2) + 1}}{n(n + 2)} = \dato{q}{1 + \frac{1}{n(n + 2)}}

    The numerator of the base exceeds the denominator by 11, because (n+1)2=n2+2n+1=n(n+2)+1(n + 1)^2 = n^2 + 2n + 1 = n(n + 2) + 1.

  3. (1+1n(n+2))n+1≥1+n+1n(n+2)\dato{B}{\left( 1 + \frac{1}{n(n + 2)} \right)^{n + 1} \geq 1 + \frac{n + 1}{n(n + 2)}}

    By Bernoulli’s inequality, with x=1n(n+2)x = \frac{1}{n(n + 2)} and exponent n+1n + 1, the power of this base is at least 1+(n+1)x1 + (n + 1)x.

  4. bnbn+1≥(1+n+1n(n+2))n+1n+2=(n2+3n+1)(n+1)n(n+2)2\frac{b_n}{b_{n + 1}} \geq \resaltar{\left( 1 + \frac{n + 1}{n(n + 2)} \right)} \frac{n + 1}{n + 2} = \resaltar{\frac{(n^2 + 3n + 1)(n + 1)}{n(n + 2)^2}}

    We substitute into the quotient and reduce to a common denominator; the sum n(n+2)+(n+1)n(n + 2) + (n + 1) equals n2+3n+1n^2 + 3n + 1.

  5. (n2+3n+1)(n+1)n(n+2)2=n3+4n2+4n+1n3+4n2+4n>1\frac{\resaltar{(n^2 + 3n + 1)(n + 1)}}{\resaltar{n(n + 2)^2}} = \frac{\resaltar{n^3 + 4n^2 + 4n + 1}}{\resaltar{n^3 + 4n^2 + 4n}} > 1

    We expand the numerator and the denominator: the numerator exceeds the denominator by 11.

  6. b1=4>b2=278>b3=25681>b4=31251024b_1 = 4 > b_2 = \frac{27}{8} > b_3 = \frac{256}{81} > b_4 = \frac{3125}{1024}

    Therefore, bn>bn+1b_n > b_{n + 1}. In the figure, the terms bnb_n go down while the terms ana_n go up, and each bnb_n lies above each ana_n.

The number e

The two sequences approach each other, but they do not cross: every term of the first is less than every term of the second, even if their indices are different.

LemmaThe two sequences do not cross

For all natural numbers nn and mm we have an<bma_n < b_m.

The argument passes through the larger of the two indices: the first sequence increases, the second decreases, and at each index the second lies above the first. The figure follows the case n=2n = 2 and m=5m = 5.

Proof

  1. bp=ap(1+1p)>ap\dato{p}{b_p = a_p \left( 1 + \frac{1}{p} \right) > a_p}

    At the same index pp, the term bpb_p is apa_p multiplied by 1+1p1 + \frac{1}{p}, which is greater than 11, and apa_p is positive.

  2. n≤p→an≤ap\dato{a}{n \leq p \rightarrow a_n \leq a_p}

    Let pp be the larger of the natural numbers nn and mm. Since n≤pn \leq p and the sequence ana_n is increasing, its term of index nn does not exceed the one of index pp.

  3. m≤p→bp≤bm\dato{b}{m \leq p \rightarrow b_p \leq b_m}

    Since m≤pm \leq p and the sequence bnb_n is decreasing, its term of index pp does not exceed the one of index mm.

  4. an≤ap<bp≤bma_n \leq a_p < b_p \leq b_m

    Combining the three inequalities, the middle one strict, gives the lemma. In the figure, with n=2n = 2 and m=5m = 5, the larger index is p=5p = 5.

It follows that the set of the terms ana_n is nonempty and bounded above, by 33 or by any bmb_m. By the completeness of the real numbers, it has a supremum, which is denoted by the letter ee.

Definition. The number ee is the supremum of the set of the terms of the sequence (1+1n)n\left( 1 + \frac{1}{n} \right)^n, for nn a natural number.
TheoremThe number e between the two sequences

For every natural number nn we have an<e<bna_n < e < b_n.

We use the definition of the supremum as the least upper bound and the monotonicity of both sequences. The figure shows the two sequences, which approach ee from opposite sides.

Proof

  1. an<bm,n,m∈N\dato{ab}{a_n < b_m}, \quad n, m \in \mathbb{N}

    By the lemma above, every term of the first sequence is less than every term of the second.

  2. (∀m∈N)(e≤bm)\dato{eb}{(\forall m \in \mathbb{N})(e \leq b_m)}

    Each bmb_m is therefore an upper bound of the set of the ana_n, and the supremum, which is the least of the upper bounds, exceeds none of them.

  3. an<an+1≤e\dato{ae}{a_n < \resaltar{a_{n + 1}} \leq \resaltar{e}}

    Since ee is an upper bound and the sequence ana_n is strictly increasing, each term is less than the one after it, which does not exceed ee.

  4. e≤bn+1<bn\dato{be}{e \leq \resaltar{b_{n + 1}} < \resaltar{b_n}}

    Likewise, since the sequence bnb_n is strictly decreasing, ee does not exceed the term after it, which is less than bnb_n.

  5. an<e<bna_n < e < b_n

    Combining both inequalities gives the theorem: ee is the only number that lies between all the ana_n and all the bnb_n.

  6. a74<e<b17a_{74} < e < b_{17}

    With a calculator one finds that a74a_{74} exceeds 2.7 and that b17b_{17} is less than 2.8; therefore, ee lies between 2.7 and 2.8. The bracketing is slow: the distance between the two sequences is bn−an=annb_n - a_n = \frac{a_n}{n}, which decreases like the reciprocal of nn.

The number ee is approximately 2.718281828. Jakob Bernoulli found it in 1683 while studying compound interest, and Leonhard Euler, who discovered a good many of its properties, denoted it by the letter ee. Together with π\pi, it is one of the two fundamental constants of analysis.