convergence of the sequence (1+1/n)^n

Theorem 1.

The following sequenceMathworldPlanetmath:

an=(1+1n)n (1)

is convergentMathworldPlanetmathPlanetmath.


The proof will be given by demonstrating that the sequence (1) is:

  1. 1.

    monotonic (increasing), that is an<an+1

  2. 2.

    bounded above, that is n,an<M for some M>0

In order to prove part 1, consider the binomial expansion for an:


Since i{1,2(k-1)}:(1-in)<(1-in+1), and since the sum an+1 has one term more than an, it is demonstrated that the sequence (1) is monotonic.
In order to prove part 2, consider again the binomial expansion:


Since k{2,3n}:1k!<12k-1 and n(n-1)(n-(k-1))nk<1:


where the formulaMathworldPlanetmathPlanetmath giving the sum of the geometric progression with ratio 1/2 has been used. ∎

In conclusionMathworldPlanetmath, we can say that the sequence (1) is convergent and its limit corresponds to the supremum of the set {an}[2,3), denoted by e, that is:


which is the definition of the Napier’s constant.

Title convergence of the sequence (1+1/n)^n
Canonical name ConvergenceOfTheSequence11nn
Date of creation 2013-03-22 17:43:26
Last modified on 2013-03-22 17:43:26
Owner kfgauss70 (18761)
Last modified by kfgauss70 (18761)
Numerical id 7
Author kfgauss70 (18761)
Entry type Theorem
Classification msc 33B99
Related topic NondecreasingSequenceWithUpperBound