Infinity+1 over infinity

Fuck this shit

He means [eqn] \lim_{n \rightarrow \infty} \mid{\dfrac{a_{n+1}}{a_n}}\mid
= \mid\dfrac{a_{\infty+1}}{a_\infty}\mid. [/eqn]

He means [eqn] \lim_{n \rightarrow \infty} \left|{\dfrac{a_{n+1}}{a_n}}\right|
= \left|{\dfrac{a_{\infty+1}}{a_\infty}}\right|. [/eqn]

always leave a space before the backslashes
Veeky Forums doesn't ALWAYS need it,
but I haven't figured out where it does need it, so fuck it, I always write " \" instead of "\"

but [math]a_\infty [/math] has no meaning in this context. The sequence is indexed by positive integers. there is no infinity in the integers.

You stinkendes kleinwüchsiges brainlet

You picked up an algebra 3 book and you don't even know what an element or a set is.

This is a bully free thread

LOL You're going to fail calc II because you're a fucking retard.

Pray tell, in the axioms of ZFC, tell me what is a set, what is an element, and what does it mean for an element to belong to a set?

Explain to me this if ur so smart

I know you are blind and retarded but the smell of your mom must have rubbed off while I was fucking her, alas I am not here to babysit you.

If you know how sets and maps work, and I assume numbers too, how the fuck are you still confused?

Is that a James Stewart Calculus (8th edition)?

Yeh lol, but it's 5th edition

I knew it was Stewart the moment I saw it. I'll never forget my first calculus book. *wipes away a single tear*

Anyways, OP. Essentially, you're a retard who doesn't understand calculus and I'm surprised you've even made it this far.

[math]\infty[/math] is not a real number. We consider the behavior of an as [math]{n\rightarrow{\infty}}[/math]. We do this by taking a limit of the general expression of [math]a_{n}[/math].

Explain in English pls

So you don't know what a limit is?

I get that. I don't understand why you can take a limit of infinity and add one to it. I'm a brainlet I know

See

I thought you were trolling. Sorry for being a dick about it.

The [math]n+1[/math] bit is actually just considering the next element in the sequence. For example, let us compute [math]\lim_{n\rightarrow{\infty}}\frac{|a_{n+1}|}{|a_{n}|}[/math] when [math]a_{n}=\frac{1}{2^{n}}[/math].

The we have

[math]\lim_{n\rightarrow{\infty}}\frac{|a_{n+1}|}{|a_{n}|}=\lim_{n\rightarrow{\infty}}(\frac{1}{2^{n+1}}})(\frac{2^{n}}{1})=\frac{1}{2}[/math].

Sorry for deleting this post before. I keep fucking up my latex.

Or maybe Veeky Forums's latex compiler is rudimentary as fuck and cannot compute even the most basic commands.

That last line is supposed to read:

lim((1/2^(n+1))(2^n)/1))=1/2. See?

You don't just substitute in whatever the limit is. Go back to the formal definition of limits.

You're not.
You're looking at the ratio of the nth term of a sequence and the nth+1 term.

The tends to infinity is the definition of a limit for convergence.

The limit exists if there's some part of the sequence such that all subsequent terms of the sequence approach a constant value to arbitrary accuracy.

If the sequence approaches a value and then at a later point approaches a different value that's bigger, then the limit of sum of the terms in the sequence diverges.

You don't substitite infinity. That's retarded. It's just a reminder you're in the natural numbers and there's no largest natural number. Unless you're wildburger.

I get what the limit means and i understand what you guys are saying. What I'm having trouble understanding is this (n+1) as n approaches infinity. For example, lets say a geometric series like (1/2)^n approaches to 0 as n approaches to infinity. What I'm confused about is if (1/2)^n+1 is also approached to infinity as n approaches to infinity, wouldn't that number exceed the bounds of the series?

"number" is what i meant

You are concerned with the limit of the RATIO of those two terms, as your picture shows. I showed that this limit goes to 1/2.

>(1/2)^n
en.wikipedia.org/wiki/1/2_+_1/4_+_1/8_+_1/16_+_⋯

That series doesn't approach 0, it approaches 1. The ratio test just asks you to compare any element of that series with the element coming after it. If the ratio of those 2 terms is less than one, then the SERIES CONVERGES. It doesn't fucking matter if the elements of the series approach a limit. what you're asking is whether the series converges or not.

That is an oddly specific wiki article...

yeah, a bit surprising, but also not considering how widely known it is. I think people use it all the time to introduce the concept of convergent series because it's easy to visualize.

Supposed to equal 1/(-1/12)

check under "see also"

en.wikipedia.org/wiki/0.999...

this thread has been primed for derailure

en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯

Neither can exceed bounds or else thered be a largest natural number, which is a contradiction to peanos axioms.

It's important to remember the sequence n+1 is a tail of the sequence n.

So what's really being said is if some tail of the sequence (n+1) approaches a value larger than the sequence contained by it (n) then the original sequence diverges, so the sum diverges.

If however, the tail is decreasing, then the sequence (n) converges so its sum converges.

If the sequence is constant, it can converge or diverge. So you don't know if the sum will converge or not.

Sorry, I meant that if the sequence is constant, then the sum can diverge.
Ex. {an} = 1 has limit 1
Then an+1/an = 1/1 = 1
But sum 1 from 1 to infinity is divergent.

damn you people are retarded. The +1 isn't in the subscript. You can't have a limit as n -> infinity+1 that doesn't even make sense.

a is a function of n, evaluate as lim n-> infinity and then afterwards add the 1.

Why are you trolling? OP is honestly interested.

The +1 is in the subscript. It means you're comparing an element in a sequence to the next element in a sequence at the limit.

If ratio is 0 then the next term tends to be higher than the previous one which means the series diverges.

If the ratio is 1 then the previous and the next terms tend to be equal which can mean the series converges (both terms really close to 0) or diverges.

use the tex previewer, brainlet, and quit projecting your failures onto Veeky Forums

It's 0.999... for binary. So yeah, it's going to have a wiki article, fifteen thousand questions on stackexchange, a subreddit, etc.

I actually typed it out in another editor I use before hand, but thanks for showing me that

Except /Sci doesn't use the standard syntax for latex, which is why I don't use it here.

bunch of shit that doesn't mean anything

Did you skip or """speedread""" the sections that covered limits?

Re-read the section on limits, retard.

convergence is so fucking stupid holy shit. In preschool they just called it estimation. Fuck off.

>dat name

Soon, it will be legal to genocide every commie in Western nations. Feelsgoodman.

does a set of all numbers contain itself+1?

I am currently taking Calc II at /uni/ (((ASU)))

I have learned to ignore the text book and professor to have pic related teach me about calculus

khanacademy.org/math/

I got a 93 on the last test.

> [math]|\frac{a_{n+1}}{a}|[/math]
why not [math]|\frac{a}{a_{n-1}}|[/math]?

I meant [math]a_n[/math] instean of [math]a[/math].

its not though

It's already been answered if you bothered to read.

an+1 is a subsequence of an.
Therefore, an+1 > an implies an diverges.
Therefore, the series diverges.

estimation is completely different from the concept of convergence

This is nonsense, there is no difference between using |a_{n+1}/a_n| and |a_n/a_{n-1}|, you're just shifting indices. Also, a_{n+1} isn't a subsequence in any meaningful sense, it's the same sequence, or rather, a term in the same sequence.

You should receive read analysis.
There's no guarantee the a_n-1 term even exists.

Also, a_n+1 is considered a subsequence of a_n as all terms of n+1 are contained in n.

Obviously, in this context, an and an+1 are sequences, not numbers.

Looks like Rogouzki (sic) single variable calculus latest edition.

I'm just gonna think and do it like that honestly

No, when performing these tests, they are numbers in the sequence. and we don't care if the first element exists because we're looking at limits, only the tails matter, which are the same. Again, I agree that it's technically a subsequence but not in any meaningful way.

thx 4 mansplaining wot he means

have you ever heard of the terms 'subscript' or 'series'?