convergence of the sequence (1+1/n)^n
Theorem 1.
Proof.
The proof will be given by demonstrating that the sequence (1) is:
-
1.
monotonic (increasing), that is an<an+1
-
2.
bounded above, that is ∀n∈ℕ,an<M for some M>0
In order to prove part 1, consider the binomial expansion for an:
an=n∑k=0(nk)1nk=n∑k=01k!nnn-1n…n-(k-1)n=n∑k=01k!(1-1n)…(1-k-1n). |
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:
an=1+nn+12!n(n-1)n2+13!n(n-1)(n-2)n3+…+1n!n(n-1)…(n-n+1)nn. |
Since ∀k∈{2,3…n}:1k!<12k-1 and n(n-1)…(n-(k-1))nk<1:
an<1+(1+12+12×2+…+12n-1)<1+(1-12n1-12)<3-12n-1<3 |
where the formula giving the sum of the geometric progression with ratio 1/2 has been used.
∎
In conclusion, 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:
lim |
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 |