If you can solve this you are Veeky Forums elite

If you can solve this you are Veeky Forums elite.

too many variables
What do you do? Find a closed form for the first equation and plug it into the second?

The sad thing is that even this is baby tier. It truly is terrible the low quality of intelligence and knowledge here on Veeky Forums.
Any real mathematician is over on math stackexchange and never even touches this place.

If it helps the answer is [math]\left(\frac{k+1}{k}\right)^k[\math]

Please explain to me how it get a solution for this baby tier problem

[math]\left(\frac{k+1}{k}\right)^k[/math]

I'd love to work on this problem more
I have homework due in an hour I have to turn in
Just go to math stack exchange and don't come back here if you want to be a real mathematician

I'm here, so clearly I'm not a real one

>Any real mathematician is over on math stackexchange and never even touches this place.

do you really need to point this out and lament the fact?

where exactly do you think you are?

Apparently a place where people revel in pretending to be intelligent and know things

user-sama, how can someone get on your level? Reading textbooks?
Pliz help
honestly need sensei

I'm a dropout who let rust accumulate over the years. However I use to be on track to be really good as a teenager.

Unfortunately I think it's largely conditioning. Even if you want to be as good as you can be, it entails immersing yourself completely in mathematics and work and, involuntarily, losing a large part of the rest of your personality at the end.

You cannot teach someone how to think. All I can say is, understand everything you come across to the deepest level you can. Never accept "I can't understand it." Bang your head on the wall for hours or days. Always push your thought process. Never be lazy.

Most of all though, don't suffer. Because that's what matters in life the most.

temperment and innate essence I mean. Not conditioning

I know everything. I have a 170 IQ but will not waste breath simply because you brainlets won't understand a word I say.

Yep that's pretty much Veeky Forums
m-muh IQ

...

Something is wrong with this problem
Assuming a_n becomes large, otherwise the limit would be zero, we see [math] a_{n+1}-a_{n} = \frac{1}{a_n^{1/k}}

What will happen is since a_n grows arbitrarily slowly, n will become arbitrarily larger than a_n.

There is no room to say the higher power of the numerator keeps it from equaling zero

Since [math]\forall x\,>\,0,\,x\,+\,\frac1{\sqrt[k] x}\,>\,x[/math] and [math]x\,=\,x\,+\,\frac1{\sqrt[k] x}[/math] has no real solution, [math]\lim_{n\,\to\,\infty} a_n\,=\,\infty[/math].

For all [math]n\,\in\,\mathbf N[/math],
[eqn]a_n\,+\,\frac1{\sqrt[k]{a_n}} \,=\, \sqrt[k]{{a_n}^k \,+\, \frac1{a_n}} \,=\, a_n\sqrt[k]{1 \,+\, \frac1{{a_n}^{k\,+\,1}}}[/eqn]

Since [math]\lim_{n\,\to\,\infty} \frac1{{a_n}^{k\,+\,1}} \,=\, 0[/math], we can do the following Taylor expansion:
[eqn]\sqrt[k]{1 \,+\, \frac1{{a_n}^{k\,+\,1}}} \,=\, 1\,+\,\frac1k \,\times\, \frac1{{a_n}^{k\,+\,1}} \,+\, \underset{n\,\to\,\infty}o \left(\frac1{{a_n}^{k \,+\, 1}}\right).[/eqn]
Thus,
[eqn]a_{n\,+\,1} \,=\, a_n\,\sqrt[k]{1 \,+\, \frac1{{a_n}^{k\,+\,1}}} \,=\, a_n\,+\,\frac1k \,\times\, \frac1{{a_n}^k} \,+\, \underset{n\,\to\,\infty}o \left(\frac1{{a_n}^k}\right).[/eqn]

At least that's what I think is the hardest part. Too lazy to finish.

nothing worse than being a bitter faggot dude

The solution of the recurrence grows with the same speed as the solution of the DE
[eqn]\dot x = \frac{1}{x^k} \\
x(0) = a_0 [/eqn]

This DE has the solution:
[eqn] x(t) = \left(\left(\frac{k+1}{k} \right) t + a_0^{\frac{k+1}{k}} \right)^\frac{k}{k+1} [/eqn]

Now it's easy to see that
[eqn] \lim_{t \to \infty} \frac{\left( x(t) \right)^{k+1}}{t^k} = \left(\frac{k+1}{k} \right)^k [/eqn]
so
[eqn] \lim_{n \to \infty} \frac{a_n^{k+1}}{n^k} = \left(\frac{k+1}{k} \right)^k [/eqn]
too.

The DE should be
[eqn]\dot x = \frac{1}{x^{\frac{1}{k}}} \\ x(0) = a_0 [/eqn]
obviously.

[math] \left( \frac{k+1}{k} \right)^k[/math]

My money is on 0

Me too

The difference between successive terms in a_n clearly becomes arbitrarily small
These people saying otherwise probably made a mistake

It’s not 0 I posted the solution

0? i mean at the bottom the the n to the k would just be infinitely large, and a subindex is somewaht apparent that world grow more slowly as n approaches infinty and as such a.n grows more and more slowly each time meaning that the numerator grows slower than the denominator.??????

the difference between a(n) and a(n+1) gets smaller and smaller approaching zero, so the sum of all the a(n)s should approach a finite # as n gets large.
so with that knowledge, approaching the second part, as n approaches infinity the numerator also approaches a finite number or at any rate grows much slower than the denominator, n^(k) which goes to infinity and the whole limit of the expression approaches zero. is that correct? i just reasoned through it in my head.

Your solution is incorrect.

>the difference between a(n) and a(n+1) gets smaller and smaller approaching zero, so the sum of all the a(n)s should approach a finite # as n gets large.
This is wrong. Consider a(n) = 1/n for an obvious counterexample.

I'm getting this.
First I write b_n = (a_n)^(k+1) / n^k. These are the terms of the limit I am concerned with.
Next I write a_n = (b_n)^(1/(k+1)) * n^(k/(k+1)).
Substitute this into the recursive eq.
After some manipulation you get (b_(n+1))^(1/(k+1)) * (1+1/n)^(k/(k+1)) = (b_n)^(1/(k+1)) * (1+((b_n)^(-1/k))/n).
(b_(n+1))^(1/(k+1)) and (b_n)^(1/(k+1)) cancel as n gets big.
(1+1/n)^(k/(k+1)) = 1+(B^(-1/k))/n.
n*((1+1/n)^(-1/(k+1)) - 1) + (1+1/n)^(-1/(k+1)) = B^(-1/k).
L'Hospital to get B^(-1/k) = -1/(k+1) +1 = k/(k+1)
B = (1+1/k)^k

OP here. Here’s a possible way to solve it.

fug

We get it, you wasted your life in math! Thanks user

>We get it, you wasted your life in math! Thanks user
Just where do you think you are, user?