(1+1/n)n is an increasing sequence


Theorem 1.

The sequence (1+1/n)n is increasing.

Proof.

To see this, rewrite 1+(1/n)=(1+n)/n and divide two consecutive terms of the sequence:

(1+1n)n(1+1n-1)n-1 = (n+1n)n(nn-1)n-1
= ((n-1)(n+1)n2)n-1n+1n
= (1-1n2)n-1(1+1n)

Since (1-x)n1-nx, we have

(1+1n)n(1+1n-1)n-1 (1-n-1n2)(1+1n)
= 1+1n3
> 1,

hence the sequence is increasing. ∎

Theorem 2.

The sequence (1+1/n)n+1 is decreasing.

Proof.

As before, rewrite 1+(1/n)=(1+n)/n and divide two consecutive terms of the sequence:

(1+1n)n+1(1+1n-1)n = (n+1n)n+1(nn-1)n
= ((n-1)(n+1)n2)nn+1n
= (1-1n2)n(1+1n)

Writing 1+1/n as 1+n/n2 and applying the inequalityMathworldPlanetmath 1+n/n2(1+1/n2)n, we obtain

(1+1n)n+1(1+1n-1)n (1-1n2)n(1+1n2)n
= (1-1n4)n
< 1,

hence the sequence is decreasing.

Theorem 3.

For all positive integers m and n, we have (1+1/m)m<(1+1/n)n+1.

Proof.

We consider three cases.

Suppose that m=n. Since n>0, we have 1/n>0 and 1<1+1/n. Hence, (1+1/n)n<(1+1/n)n+1.

Suppose that m<n. By the previous case, (1+1/n)n<(1+1/n)n+1. By theorem 1, (1+1/m)m<(1+1/n)n. Combining, (1+1/m)m<(1+1/n)n+1.

Suppose that m>n. By the first case, (1+1/m)m<(1+1/m)m+1 By theorem 2, (1+1/m)m+1<(1+1/n)n+1. Combining, (1+1/m)m<(1+1/n)n+1. ∎

Title (1+1/n)n is an increasing sequence
Canonical name 11nnIsAnIncreasingSequence
Date of creation 2013-03-22 15:48:39
Last modified on 2013-03-22 15:48:39
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 14
Author rspuzio (6075)
Entry type Theorem
Classification msc 33B99