Prove me wrong

Prove me wrong.

Pro tip: You can't.

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[eqn] \sum_{n=1}^\infty q^n = \frac{q}{1-q} [/eqn]
Let [math]q = 2 [/math]
[eqn] \sum_{n=1}^\infty 2^n = \frac{2}{1-2} = -2 [/eqn]

lim [n-> infinity] (n/n*2^n) = infinity

The expression being summed is strictly positive for all n >= 1. This implies that the sequence of partial sums is increasing. If the sequence doesn't converge, then your statement is false. If the sequence of partial sums converges, then the limit must be >= the first term of the sequence, since the numbers being summed are strictly positive. Since the first term in the sequence of partial sums is 2, the limit, if it exists, is >=2.
Neither of the two allowable options concludes that the limit of the sequence of partial sums is 1, so I conclude that you are wrong.

The equation you are using for the value of an infinite geometric sum only applies when -1 < q < 1, otherwise the series diverges, user.

I think user was joking. It made me giggle

lmao

you complete brainlet, geometric series converges ONLY for [math] \lVert q \rVert < 1[/math]

t.
just took linear algebra for brainlets

wut?

most intelligent post on Veeky Forums

what if the limit was 3?

Because of the uncertainty principle, there's no way to tell. Limits greater than 2 are isomophic to the naturals, so they are symmetric and indistinguishable. Heisenberg discovered this like a million years ago, come on

Fucking Calc II teacher spent all the time speaking about limits and not a word about Heisenberg. I blame the education system.

[math]\sum_{n=1}^k 2^n\geq \sum_{n=1}^k 1^n=k\geq 0[/math] and [math]k\to\infty[/math]. Hence [math]\sum_{n=1}^{\infty} 2^n[/math] diverges

why the stupid fucking n/n

n = 2
2^2 = 4
any number of positive integers + 4 > 1
S_n > 1

>any number of positive integers + 4 > 1
prove it

that's actually directly given by the construction of the naturals from Peano brainlet

It's actually -2 OP:
en.m.wikipedia.org/wiki/1_+_2_+_4_+_8_+_⋯

It equals -2 but doesn't converge to -2.

...

Fun fact: It does converge to -2 in the 2-adic numbers.

>not wrapping the parentheses
Am I the only one really bothered by this?

>why the stupid fucking comment, fgt pls

This but unironically

>only introduced to limits in calc ii
let me guess, calc i was """""""""college algebra""""""""

hes referring to your usage of the norm

Either almost every post in this thread is trolling or I am really lost here

I'm a little rusty on my algebra, but seem to remember learning somewhere around age 5 that [math]\frac{n}{n} = 1[/math], which would mean that your sum is actually

[math]\sum_{n=1}^{\infty} 2^n[/math]

And clearly, that sum adds to [math]-\frac{1}{12}[/math]

kek

Clever trick to create new identeties
[eqn]1=\sum_{n=0}^\infty \frac{n\cdot2^n}{n}=\frac{\sum_{n=0}^\infty n\cdot2^n}{\sum_{n=0}^\infty n}=\frac{\sum_{n=0}^\infty n\cdot2^n}{\frac{-1}{12}}\implies -12=\sum_{n=0}^\infty n\cdot2^n[/eqn]

this is because wolfram is a jew and is trying to hide the fact that by paying his customers in increasing increments of $2, he is actually scamming them out of $2.

[math]2^x[/math] is the more natural and better exponential then [math]e^x[/math].

Proof:

[eqn]2^x = \sum^{\infty}_{k = 0}\frac{1}{k!} \Gamma(x + k)/\Gamma(x) [/eqn]