How to prove monotone sequences

in this video I used Bernoulli's inequality to show that a sequence in monotonic increasing. I took this approach because not all classes allow the use of derivatives to show that a sequence in monotone

[HenkVanLeeuwen-io]

For all y<e there exists x so if n>x then y<a(n) <e. So limit of a(n) =e.

[sobolzeev]

The Bernoulli inequality actually is
(1 + x)ⁿ > 1+ nx for x≥-1, x≠0, n≥2. So the sequence is strictly increasing.

Proof by induction.
For the base n=2, this is
1 + 2x + x² > 1 + 2x.

For the step of induction, since (1+x)≥0,
(1+x)ⁿ⁺¹ ≥ (1+nx)(1+x)
= 1 + (n+1)x + nx²
> 1 + (n+1)x.

[holyshit]

As a bonus show that this limit is bounded
and we will be able to conclude that this sequence is convergent

[sobolzeev]

By the way, there is a fourth way to prove it: by contradiction. If the sequence is NOT monotone increasing while we definitely have a₁<a₂, then there exists an index n such that aₙ₋₁<aₙ ≥ aₙ₊₁. That is,
aₙ² > aₙ₊₁ aₙ₋₁.
However, we still need the Bernoulli inequality to disprove this.

[lawrencejelsma]

Bernoulli's approximation helps figure out sequences like an and an+1 to see if it is increasing for an arbitrary n. You explained using the >= 1 + nx so clear and clean in your proof that I'll find whenever I have to check a sequence by Bernoulli's Principle I can also see if an n+1 term is still greater than a n term result. 👍

[AbouTaim-Lille]

In analysis every injective continuous function over a given interval [a, b] is monotonic.
The proof is simple.
For simplicity we can suppose that f(b) ≥ f(a) and prove that f is increasing.
Let's suppose that f is Not increasing. Then there are 2 points X and y where a<X<y<b and f(X)>f(y) but since f is a real continuous function it admits any value ζ in the interval [f(y), f(X)].
Now let's put ζ = max {a, f(y)}
And by continuity there is a point
X1 in the interval [a, X] such that f(X1) = ζ, similarly there is a point
X2 in the interval [X, y] such that f(X2) = ζ
Thus we have two points X1 and X2 satisfying f(X1) = f(X2) whereas x1≠x2 which contradicts with the fact that f is injective.

[richardheiville]

In fact the inequality you use is (1-x)^n>1-nx, x<1

[barryzeeberg]

0:52 it is not obvious that the sequence is increasing, the only thing that is obvious is that the first several terms are increasing, anything can happen after that . . .

[Orillians]

Whats an example of non-monotone sequence.

[lornacy]

In the sixth step you casually replaced the 1 in the numerator with n. It's not clear to me why you can do that. Will someone explain? ☺️

[boguslawszostak]

The derivative of f(x) must be greater than or equal to 0.

[MegoElazab]

How to prove that equal e when n tends to ifinity