Prove that this equation is wrong

That's obviously wrong because it's 1/2 + some other positive stuff.

if a series doesn't converge it's divergent full stop

Wtf is the retardation in this thread.
The sum in ops picture is convergent (to 1)

your the retard you need to work on you're reading comprehension

Zenos paradox is itself a paradox. In his achilles and tortoise example he maps multiple factors that allow for a finite result, and in practice use finite inputs as well. The tortoise gets a head start but moves at half the speed of achilles. When achilles reaches where the tortoise started at 0.5 the race track, the tortoise has made it to 0.75. When achilles reaches 0.75, the tortoise has teached 0.875, so on until achilles reaches the tortoise.

The multiple factors are time, distance, and a surmised end or point at why achilles and the tortoise meet. In fact the graph for the ewuation shows divergent values above 1. Taking altogether, it's just retard bullshit to take it as proof of infinite work sums. Taken without the time or limit distance factor, there would always be a smaller slice of the line between achilles and the tortoise, therefore never reaching 1. Taken with the time and limit distance factors, they reach 1 at some moment and then achilles diverges by doubling the distance between himself and the tortoise behind him with every reference frame, 1 to 2 to 4 to 8 to 16 to 32 to 64 to 128 and so on until achilles as run off the face of the earth and travelled across the universe towards infinite distance.

Zeno was a fucking brainlet and everyone who treats him other than the fact will also be a brainlet. Infinite sum is stupid gay stuff.

Daily reminder that a discrete spacetime solves Zeno's paradox.

When achilles reaches where the tortoise started at 0.5 the race track, the tortoise has made it to 0.75. When achilles reaches 0.75, the tortoise has teached 0.875, so on until achilles reaches the tortoise.
> there would always be a smaller slice of the line between achilles and the tortoise
> never reaching 1

[math]0.111..._2=1[/math]

Daily reminder that Calc I solves Zeno's paradox.

This just means that the application of infinite sums only works for finite things, but even then it would have to be extended to either defining or getting as a result the actual limit point at which a convergence occurs.
Applied arbatrarily to numbers and variables without finite limits is not the correct way to use it. For example it's easy to mechanically prove [math]\sum_{n=1}^{\infty} \frac{9}{10^{n}} = 1[/math] if and only if we define a finite decimal accuracy limit, which as a product would also define the n limit at which the decimal limit was reached, which for this problem would just be decimals+1. For 10 decimals of accuracy, on the 11th iteration it would round up to 1 from 0.999999999 at the 10th, so truthfully
[math]\sum_{n=1}^{X} \frac{9}{10^{n}} = 1 with 10 decimal accuracy[/math], but written more simply lacking necessary information would instead result
[math]\sum_{n=1}^{\infty} \frac{9}{10^{n}} = 0.\bar{9}[/math]

Yes, the sort of 'stupid gay stuff' that shares itself with one other infinity descriptor: "I'm not stupid or gay enough to think of a way to ultimately not be a super-immortal quasi-dimensionality."

It's like being Christian and seeing the sign for that whole 'shared belief' concept.

oh god it's the shitlatex masturbator again

>of course the mouth breather can't spell

m.wolframalpha.com/input/?i=N[sum[1/(2^n), {n, 1, 118}], 35]
This would suffice to solve achilles and the turtle if the smallest measurement rererence frame was a planck length with 35 decimals of accuracy. At n=118, 1/2^n converges to 1 with planck length accuracy. At n=117, its just 0.999...

>just going around proving zeno's paradoxes wrong on a chinese pottery bulletin board in the middle of the night on new years

fuck you it's noon here

>只是繞來繞去證明芝諾悖論錯在中國陶公告牌在新年內半夜

*SLAM DUNK*

but
[math]\frac{1}{2^n}+1 \neq \frac{1}{2^n}+\frac{1}{2}[/math]

...

...

Isn't that what I said? You can't use induction on this shit anyway, can you?

It means you should go back to your board

holy

Name one (1) intelligent Republican.

I'm conservative republican and have proven one of zeno's paradox solutions/assumptions incorrect.

now THIS is real deal.
[eqn]
\sum\limits_{n=1}^ \infty \frac{1}{\frac{1}{n}} = -\frac{1}{12}
[/eqn]

that equation tends towards infinity

...

but the value is -1/4pi

No you cant.

No, the value is a positive number that incrementally adds up.

[math] \sum_{n=0}^m x^n = \dfrac{1-x^{n+1}}{1-x} [/math]

so for |x|

>finite operations are the same are "infinite operations"
aww cute

m.wolframalpha.com/input/?i=sum[1/(1/n), {n,1,infinity}]

n:1 = 1/(1/1) = 1 +
n:2 = 1/(1/2) = 2 + 1
n:3 = 1/(1/3) = 3.3 + 2 + 1
n:4 = 1/(1/4) = 4 + 3.3 + 2 + 1
...
[math]\rightarrow \infty[/math]

limit =/= value

take your shit in street pajeet ideologies elsewhere please. We're trying to do math here.

[eqn]S = \sum_{n=1}^{\infty} \frac{1}{2^n}=\frac{1}{2}+\frac{1}{2}S \iff S = 1[/eqn]
QED
Also algebra is universal and Cauchy was a faggot and a /pol/tard.

Are you retarded?

I Know im late but to anyone still wondering how the original result comes to be here goes...

are you?

adding to that, i don't understand why its a paradox.... i didn't even knew zeno existed we just had that series in a calc homework

t. I Don't Understand [math]\mathbb{R}[/math]

something doesn't seem about this. you give the square an area of 1 before filling in the triangles, so it's like you're starting with the solution you're trying to prove

the reasoning is that you can fit all the triangles in the unit area square I think.

Its a good representation of the problem described by zeno as he compared multiple finitist factors like time and distance resulting in a finite answer of 1, but both his problem and that image arent actually [math]\sum_{n=1}^{\infty}\frac{1}{2^{n}}[/math]
Since neither require "infinite" work to reach the result Where also describes that the least limit required to solve the problem is the amount of decimal places + 1, which if n=infinity, the decimal accuracy would have to be infinity-1, which in any practical usage would simply define infinite work and summation is not actually being done.

Required to solve a similiar problem*

>infinity is in R
o boy

If n=[math]\infty[/math] and it results 1 then how is it wrong

It's wrong by misunderstanding the usage of infinite sum. It's only 1 at n=infinity IF a decimal limit of infinity-1 was provided. Decimal limits are important because its how subdivision into smaller and smaller units work. If the decimal limit for the sum 9/(10^n) were 35 places corresponding with the planck length being [math]1.616229 × 10^{-35}[/math], n needs only reach 118 to be considered converging. Just over a hundred iterations, not a thousand or a million or a billion nevermind infinity.

Applied truly arbitrarily on numbers defined without a decimal or accuracy limit, the result should instead be nothing but [math]0.\overline{9}[/math] if we take the overline to define infinite arbitrary repitition as the result of an infinite arbitrary sum, since without a definition of accuracy limit with the only direction being [math]n\rightarrow\infty[/math] means no concatination occurs. If instead our smallest measure of time and distance was 1/10th, the sum 9/(10^n) = 1 at n=2 since 0.9 would be followed by 0.99 which would appear to be a huge chunk of a divisional increment we can't fathom, so that second 9 rounds up to 10, the first 9+1 = 0, resulting 1.0

The usage of [math]\sum_{n=1}[/math] is taught and used poorly for lack of understanding the limit is either something you should be supplying to find a decimal accuracy limit, or solving as a variable when supplied with a decimal accuracy limit.

>the limit is either something you should be given to solve a decimal accuracy limit, or solving as a variable when supplied with a decimal accuracy limit.
fixed

[math]\sum_{n=1}^{\stackrel{x}{d.10^{-35}}\frac{9}{10^{n}}[/math]
or
[math]\sum_{n=1}^{\stackrel{472}{d.x}}\frac{9}{10^{n}}[/math]
would both work as example ways to write the questions if the intent was to produce an answer indicative of convergence.

[math]\sum_{n=1}^{\stackrel{472}{d.x}}\frac{9}{10^{n}}[/math] or
like
[math]\sum_{n=1}^{\stackrel{x}{d.35}}\frac{9}{10^{n}}[/math]

>wall of text
why do all religious nuts have verbal diarrhea

>my cant unambled to read well so u r dum

oh look you missed a bus again

Might have handled the stackrel fraction but the problem is Veeky Forums's spacing. One sum followed by another in seperate math/math blocks without text after the first block as well as a line break will mess it up.

Spoken like a true engineer who doesn't understand mathematical at all.

Daily reminder that Zeno's paradox isn't a paradox at all if the arrow lands at Zeno's foot.

>look up zenos arrow
>"motion is impossible because time is just countless instances of freeze frames"
goddamn what the hell was this guy's problem?

His brain was two sizes too small.

I never said that, brainlet. Infinity is not where the other user failed to grasp the set of reals.
You don't need infinity to deal with limits. That's what that whole epsilon-delta thing is for. (=

It's a visual representiation of the fact that:

Given any ε>0, there exists a halving-step N such that after it (n>N) the difference between the filled area and the area of the square(which is 1) is less that ε.

That's what [math] \sum\limits_{i=1}^{\infty}\frac{1}{2^i} = 1 [/math] means:
[math] \forall \varepsilon >0 : \exists N : \forall n : n>N \implies |\sum\limits_{i=1}^{n}\frac{1}{2^i} - 1|

Spoken like a loser who failed Calc I.

Not sure but it looks like the only practical usage of epsilon delta is for finding the zero of a wave function. Or working to redundantly prove a given non-infinite limit alongside a given result and finding the solution for the limit via epsilons and deltas which doesn't translate to hard numbers.

Its much easier to produce results through sigma sums if you just provide the sense and scale of required outputs via non-infinite limit or non-infinite decimal accuracy. In the case of 1/2^n with n to infinity, the answer would never even equal [math]0.\overline{9}[/math] because the summation always adds on an arbitrary halved amount of various whole and different integers, unlike for example 9/10^n which produces nothing but more 9's.

There is a very slipperly slope of brainlet logic derived from shit education surrounding these problems. First its [math]0.\overline{9} = 1[/math], then it's [math]0.999........99764243367532 = 0.\overline{9} = 1[/math], and soon enough it'll just be [math]3=\infty[/math]

Draw the line somewhere bitte.

It's funny how people expect real numbers to behave like real stuff, ignoring the absurdity of their definitions. I blame primary teachers.

I believe that not even you are understanding what you are saying desu.

(You)'s don't mean as much when you don't work for them. Put a little effort in next time.

You are both correct. "The series diverges" and "the series equals -1/12" are not contradictory, just as "the alternating harmonic series does not converge absolutely" and "the series converges to ln(2)" are not contradictory. You're just extending the concept of a series in a unique way.

[math]\sum_{n=1}^{n \rightarrow \infty} \frac{1}{2^{n}} < 0.\bar{9} < 1[/math].
.
.
[math]\sum_{n=1}^{n \rightarrow \infty} \frac{9}{10^{n}} = 0.\bar{9} \neq 1[/math].
.
.
[math]\sum_{n=1}^{\stackrel{n \rightarrow x}{d.35}} \frac{9}{10^{n}} = 1, x=118 [/math].
.
.
[math]\sum_{n=1}^{\stackrel{n \rightarrow 118}{d.x}} \frac{9}{10^{n}} = 1, x=35 [/math].
.
.
[math]\sum_{n=1}^{\stackrel{n \rightarrow x}{d.1}} \frac{9}{10^{n}} = 1, x=2[/math]

where d is a placeholder for decimal point accuracy cause latex was having trouble displaying exponents in the small stackrel field but i'll try one more time:

[math]\sum_{n=1}^{n \rightarrow 118}{10^{-35}} \frac{9}{10^{n}} = 1[/math]

[math]\sum_{n=1}^{\stackrel{n \rightarrow 118}{10^{-35}} \frac{9}{10^{n}} = 1[/math]

Infinity doesn't exist. Done.

[math]\sum_{n=1}^{\stackrel{n \rightarrow 118}{10^{-35}}} \frac{9}{10^{n}} = 1[/math]

Ye that looks like ass. d dot works better.

The sheer amount of retardation here is unbelievable. There are maybe 3 people here who understand math and the rest are brainlets.

[math]\sum_{n=1}^{\stackrel{10^{-35}}{n \rightarrow 118}} \frac{9}{10^{n}} = 1[/math]

Looks better

This is my board

The sum of e^n from 8 to 1 is -negrative e^(8+1)

=e^9

It would be negtive e^7 because its acturally e^(8-1)

So it would look like

1 E n=8 of e^n = -e^7

>plank length is [math]10^{-35}[/math] metres as the smallest unit where things matter
>0.000000000000000000000000000000000001m
>planck time is [math]10^{-44}[/math] seconds as the the time it takes the speed of light to travel a plank length
>0.00000000000000000000000000000000000000000001s
>[math]\frac{0.00000000000000000000000000000000001}{0.00000000000000000000000000000000000000000001} = \frac{1pt}{1000000000pl} = \frac{0.0000000001pt}{1pl}[/math]
am i retarded or is this retarded

dude nice trips

No because the speed of light is a given whereas planks constant is imagninary

I don't accept that numbers exist so axiomatically your argument is invalid.

Ok, but how is this relevant?

it's not ok and isn't relevant

and you can add an infinite number of rational numbers together and get an irrational at the end. All of the partial sums are rational since the rationals are closed under addition but putting infinity in makes everything go to shit.

so what I'm saying is your post is wrong

>the axioms are hard, get rid of them

Nah, keep your street shitting pajeet math to yourself, you brainlet subhuman.

That yoy can find the exact finite result of a sigma sum via a sigma sum without unrelated delta epsilon fudged retard limit stuff by plugging in one more factor is what makes it relevant.
There is no such thing as infinite work leading to a rational number, and you only need do some finite work to reach any desired result.