Exponential Series & Euler's Number e: Formula, Derivation & Examples (Class 11 Maths)

Q: What is the value of e in mathematics?

Euler's number e is defined as the sum of the infinite series 1 + 1/1! + 1/2! + 1/3! + ..., and its value can be proved to lie between 2 and 3; more precisely, e ≈ 2.71828.

Q: What is the general formula for the exponential series?

The exponential series is eˣ = 1 + x/1! + x²/2! + x³/3! + ... + xⁿ/n! + ..., valid for every real value of x. Substituting x = 1 gives the series that defines e itself.

Q: How do you prove that e is less than 3?

By comparing the series 1/3! + 1/4! + ... term by term against the geometric series 1/2² + 1/2³ + ..., and showing 1/n! 2. Since the geometric series sums to 1/2, adding the first three terms of e's series (5/2) gives an upper bound of 3.

Every student encounters π early in geometry, tied to circles. Section A.1.4 introduces its calculus counterpart: the number e, first defined by Leonhard Euler in his 1748 calculus text. Unlike π, which arises from a geometric ratio, e is defined directly as the sum of an infinite series. This post explains how e is defined, proves its value lies between 2 and 3, introduces the general exponential series for eˣ, and works through two NCERT examples.

Defining the Number e as an Infinite Series

Consider the infinite series of numbers:

1 + 1/1! + 1/2! + 1/3! + 1/4! + …

The sum of this series is, by definition, the number e. Since factorials grow extremely fast, each term shrinks quickly, which is what allows this infinite series to settle to a finite sum in the first place — the same underlying principle seen in the infinite geometric series covered earlier in this appendix.

Proving That 2 < e < 3

NCERT proves the value lies in a specific range using a comparison technique. Since every term of the series is positive, e > 2 follows immediately from the first three terms (1 + 1 + 1/2 = 2.5). The more interesting bound is the upper limit.

Consider the two sums:

1/3! + 1/4! + 1/5! + … + 1/n! + … and 1/2² + 1/2³ + 1/2⁴ + … + 1/2ⁿ⁻¹ + …

Comparing term by term: 1/3! = 1/6 and 1/2² = 1/4, so 1/3! < 1/2²; similarly 1/4! = 1/24 < 1/2³ = 1/8, and 1/5! = 1/120 < 1/2⁴ = 1/16. By analogy, 1/n! < 1/2ⁿ⁻¹ for every n > 2. This means the entire sum (2) is less than the entire sum (3). The second sum is itself a geometric series with a = 1/4 and r = 1/2, which by the sum-to-infinity formula from the previous section equals (1/4)/(1 - 1/2) = 1/2.

Adding 1 + 1/1! + 1/2! = 1 + 1 + 1/2 = 5/2 to both sides of the inequality, the left-hand side becomes exactly the series defining e, while the right-hand side becomes 5/2 + 1/2 = 3. This proves e < 3, and combined with e > 2, gives 2 < e < 3.

The General Exponential Series for eˣ

Once e is defined, NCERT extends the idea to define an exponential series in a variable x:

eˣ = 1 + x/1! + x²/2! + x³/3! + … + xⁿ/n! + …

Putting x = 1 recovers the original series that defines e. This general formula is the workhorse for the two worked examples that follow — both involve substituting an expression in place of x and extracting a specific coefficient.

Worked Example 1: Coefficient of x² in e²ˣ⁺³

Problem. Find the coefficient of x² in the expansion of e²ˣ⁺³ as a series in powers of x.

Solution (elegant method). Write e²ˣ⁺³ = e³ · e²ˣ, splitting the constant part from the variable part. Now expand e²ˣ using the exponential series with x replaced by 2x:

e²ˣ = 1 + 2x/1! + (2x)²/2! + (2x)³/3! + …

The coefficient of x² inside this expansion is 2²/2! = 4/2 = 2. Multiplying by the constant factor e³ pulled out earlier gives the coefficient of x² in the full expression:

Coefficient of x² = e³ · (2²/2!) = 2e³

NCERT also demonstrates a longer, direct method — expanding (2x + 3)ⁿ/n! using the ordinary Binomial Theorem for each n, isolating the x² term, and summing the resulting series over all n from 2 onward. That approach also arrives at 2e³, but the factoring method e³ · e²ˣ shown above is far faster and is the preferred technique for this type of problem.

Worked Example 2: Estimating the Value of e²

Problem. Find the value of e², rounded off to one decimal place.

Solution. Substituting x = 2 into the exponential series:

e² = 1 + 2/1! + 2²/2! + 2³/3! + 2⁴/4! + 2⁵/5! + 2⁶/6! + …

Computing the first seven terms gives 1 + 2 + 2 + 4/3 + 2/3 + 4/15 + 4/45 + …, which totals at least 7.355. For an upper bound, NCERT groups the remaining terms (from the 2⁵/5! term onward) into a geometric-style bound using the ratio 2/6 = 1/3, arriving at an upper limit of 7 + 4/15 × [1/(1 - 1/3)] = 7 + 2/5 = 7.4.

Since e² lies between 7.355 and 7.4, its value rounded to one decimal place is 7.4. This same squeeze technique — bounding a series from below with a partial sum and from above using a geometric comparison — is the same method used earlier to prove 2 < e < 3.

Why This Matters Beyond Class 11

The number e reappears constantly in higher mathematics — in calculus, compound interest formulas, probability distributions, and growth/decay models. Section A.1.5 of this appendix, on the logarithmic series, builds directly on the exponential series covered here, since loge is the inverse of the exponential function.

Master the Exponential Series With ChampionsPrep

The exponential series bridges Class 11 Maths and the calculus that follows — and it shows up regularly in CUET, IPMAT, and JIPMAT quantitative sections. ChampionsPrep offers structured, NCERT-aligned practice for CBSE Class XI–XII Commerce students on exactly this kind of topic. Registration is completely free, and you only pay as you use the platform — start practising the exponential series today.

Test Your Knowledge

Q1.How is the number e defined in this post?

As the ratio of a circle’s circumference to its diameterAs the sum 1 + 1/1! + 1/2! + 1/3! + …As the square root of 3As the limit of 1/n

Q2.Between which two integers does e lie?

1 and 23 and 42 and 30 and 1

Q3.What is the exponential series for eˣ?

1 + x + x² + x³ + …1 + x/1! + x²/2! + x³/3! + …x + x²/2 + x³/3 + …1 − x + x² − x³ + …

Q4.What is the coefficient of x² in the expansion of e²ˣ⁺³?

2e³4e³e³e²

Q5.What is e² rounded to one decimal place?

7.37.47.57.0

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Frequently Asked Questions

What is the value of e in mathematics? +

Euler's number e is defined as the sum of the infinite series 1 + 1/1! + 1/2! + 1/3! + ..., and its value can be proved to lie between 2 and 3; more precisely, e ≈ 2.71828.

What is the general formula for the exponential series? +

The exponential series is eˣ = 1 + x/1! + x²/2! + x³/3! + ... + xⁿ/n! + ..., valid for every real value of x. Substituting x = 1 gives the series that defines e itself.

How do you prove that e is less than 3? +

By comparing the series 1/3! + 1/4! + ... term by term against the geometric series 1/2² + 1/2³ + ..., and showing 1/n! < 1/2ⁿ⁻¹ for n > 2. Since the geometric series sums to 1/2, adding the first three terms of e's series (5/2) gives an upper bound of 3.

How do you find the coefficient of x² in e²ˣ⁺³? +

Split e²ˣ⁺³ as e³ · e²ˣ, then expand e²ˣ using the exponential series with 2x in place of x. The coefficient of x² in that expansion is 2²/2!, so the overall coefficient is e³ · 2²/2! = 2e³.

Why is the number e important beyond this chapter? +

The number e is the base of natural logarithms, appears throughout calculus as the function equal to its own derivative, and is fundamental to compound interest, exponential growth and decay, and probability theory — making the exponential series one of the most widely used tools in higher mathematics.