Infinity+1 over infinity

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?