How to prove monotone sequences

[@HenkVanLeeuwen-i2o]

For all yx then y

[@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

[@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.

[@richardheiville937]

Newton's binomial gives a direct proof. (1+x)^n=1+nx+something positive if x>=0

[@holyshit922]

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

[@Christian_Martel]

My favourite quasi-monotonic function is f(x) = x^3. It’s increasing almost all the time, except when turning the corner at zero. 😂

[@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

[@lawrencejelsma8118]

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. 👍

[@robertveith6383]

Write a(n) and a(n + 1) or a_n and a_(n + 1), instead of what you wrote.

[@lawrencejelsma8118]

@@robertveith6383 ... There is no technical English classes agreeing to your statements in math. Keep going to school in the higher math courses. I doubt you will be using parentheses like what you propose on your homework, tests and understanding what professors write on chalkboards and class viewing screens.

[@gopikayala6551]

Cool and experienced

[@JosefinaMwanyana]

How can I show it if I have calculus

[@richardheiville937]

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

[@Orillians]

Whats an example of non-monotone sequence.

[@PrimeNewtons]

Sequence of cosines

[@Orillians]

@@PrimeNewtons Do you mean taylor series expansion or just a normal cosine sequence?

[@PrimeNewtons]

@@Orillians cos 1, cos 2, cos 3, .........

[@boguslawszostak1784]

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

[@tulsaken2754]

Can you do this derivative?

[@boguslawszostak1784]

@@tulsaken2754 It's not about that function. This is the rule that a differentiable function is increasing when its derivative is non-negative, meaning greater than or equal to zero. Back to this function The derivative of the function f(x) = g(x)^h(x) is calculated using the identity g(x)^h(x) = e^(ln(g(x)^h(x))) = e^(h(x)*ln(g(x))), and by applying the chain rule to e^z function. (e^z)'=z'*e^z, z=h(x)*ln(g(x))

[@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? ☺️

[@PrimeNewtons]

Multiply by n using Bernoulli's inequality

[@lornacy]

Oh, shoot, now I get it. The whole rational expression is the "x" and you're making it look like the x term in the right hand side of Bernoulli's inequality. I feel foolish. Luckily it's a familiar feeling for me.

[@MegoElazab]

How to prove that equal e when n tends to ifinity

[@TheFrewah]

This equation defines e

[@richardheiville937]

(1+1/n)^n=exp(n*log(1+1/n) and you can prove that limit n*log(1+1/n)=1 Easy way; n*log(1+1/n)=(log(1+1/n)-log(1+0))/1/n so the limit is the derivative of x->log(1+x) evaluated in x=0.

[@barryzeeberg3672]

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 . . .

[@gileadedetogni9054]

He said this :)

[@mil9102]

Longest intro ever?

[@PrimeNewtons]

Almost. This is not the longest.

[@comdo777]

asnwer=1x