Should be a piece of cake to be honest

Yes.

Lol that makes no sense

its a meme

Both are divergent, it does not make sence comparing them

There a clear logical process for arriving at it. It is a value that is relevant and usable. Why do you think its a meme?

[eqn] \sum_{n=0}^\infty (-1)^n = \frac{1}{2}[/eqn]

>Why do you think its a meme?
Because it is completely and utterly wrong.

I stopped caring about math when I was introduced to the concept of Ramanujan summation. What a crock of shit. If your equation can only be solved by inventing values for summations that can't converge, like some kind of math deity , then you are fucking wrong and the math is flawed. Same for algebra solutions that basically say "the correct answer is whatever the correct answer is". Thats what the math said transcribed to words but god forbid if i wrote in down in english instead of the ancient math runes the teacher word mark me wrong.
Math is logical and numbers never lie my ass. Math is just as flawed as any other human construct.

lhs = infinity = rhs
q.e.d

>human construct
WRONG!

[math]\displaystyle \sum_{n=1}^\infty \frac{n(n+1)}{2} = 1+3+6+10+15+\dots[/math]

1+3+6+10+15+..."="1+2+3+4+...

Oh I get it. The meme is that you're retarded.

Now prove that the harmonic series still diverges when you throw out all terms not containing some fixed digit d

user its the sum of everything up the that nth term.

1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 =10

Both sides diverge you utter moron.

Learn the basics of analysis before you talk.

>containing some fixed digit d
what does that even mean?

...

Say d=1. Then you have 1/1, 1/10, 1/11, 1/12, 1/13, ... 1/21, 1/31, ...1/100, ... as the terms of the underlying sequence.
Show the series diverges for d from 1 to 9

Follows from the convergence of the Kempner series. en.wikipedia.org/wiki/Kempner_series
Also that's a cute Tadakuni.

If it converges for d=1 then it converges for all d>1 as well, because for any term in the sum, replacing every occurrence of 1 with d strictly decreases it

So it suffices for us to prove divergence for d=1

>Follows from the convergence of the Kempner series.
It doesn't "follow from," the statements are clearly equivalent.
But give a direct proof

Tadakuni best girl

Of the d-digit numbers there are 9^d that don't contain any 1, and they are all greater than 10^d so the sum of their reciprocals is bounded by the sum of 1/(10)^d with itself 9^d times, i.e. (9/10)^d
The sum from d=0 to infinity of this is a geometric series and converges to a finite number, so the sum of the remaining terms must be infinite.

>the statements are clearly equivalent
They aren't. The series you posted and the harmonic series both diverge, but that does not mean that their "difference" (the Kempner series) converges.
The divergence of both [math]\sum_{n=1}^\infty \frac{1}{n} [/math] and [math]\sum_{n=1}^\infty \frac{1}{2n} [/math] does not mean that [math]\sum_{n=1}^\infty \frac{1}{2n-1} [/math] converges, for example
>But give a direct proof
I did.
>Tadakuni best girl
Obviously.

And I just realized I got the contrapositive the wrong way around in , though the same argument in works for any other digit besides 1

Also I should really start looking for letters other than 'd' for denoting arbitrary digits

have you ever heard of Taylor series?

This doesn't prove E is a subset of O
At n=1, you have 1/2 + 1 = 1

Yes?

That still doesnt break down the basics of analysis.

if f(x) = 1 + x + x^2 + x^3
and g(x) is the taylor polynomial of f(x)
then it is trivial to show that g(x) = f(x)
the key difference is that f(x) only converges when |x| < 1 and g(x) converges for any x

once you stop being a pussy you'll accept that 1-1+1-1+1... = 1/2 and the rest of the math that leads to 1+2+3+4+5...= -1/12
You're like those contrarians who insisted that proofs containing [math]\sqrt{-1}[/math] were "wrong" somehow.
Maybe you should go back to counting sticks and beans with Wildberger

>f(x) only converges when |x| < 1 and g(x) converges for any x
Are you retarded. That is just plain wrong.

>once you stop being a pussy you'll accept that 1-1+1-1+1... = 1/2
No. This violets A LOT of the most important theorems of analysis. (in R convergence is equivalent to it being a cauchy series)
It even breaks L^p spaces.

>You're like those contrarians who insisted that proofs containing −1−−−√−1 were "wrong" somehow.
No.
I am just not denying the whole field of analysis.

>Maybe you should go back to counting sticks and beans with Wildberger
Maybe have a look at some real analysis

What shit tier school do you go to that doesn't teach you how to expand the domain of a function?

Both series diverge, there's nothing to prove

>What shit tier school do you go to that doesn't teach you how to expand the domain of a function?
???
Nothing you say makes really any sense. You talk about "convergence of a function" which simply does not exist.
Convergence is about having a SERIES.

Then you go on stating blatantly false claims, such as that a non cauchy series can converge in R.

And after that you talk about something completely unrelated again.
Analytic continuation DOES NOT IMPLY THAT NON CAUCHY SERIES CAN CONVERGE OR THAT ANY OTHER PARTS OF REAL ANALYSIS ARE COMPLETELY BROKEN.

Do you even know what a metric space is? Or its completion? You should probably read up on that before you talk about stuff you dont understand.

>Nothing you say makes really any sense. You talk about "convergence of a function" which simply does not exist.
>Convergence is about having a SERIES
The function IS the series
If the series converges the function is defined, or rather, the function converges
Do you even know why [math]f(x)=\sum _{n=0}^{\infty}x^n[/math] is only canonically defined for |x| < 1 whereas [math]g(x) = \sum _{n=0}^{\infty}\frac{f^{\left(n\right)}\left(0\right)x^n}{n!}[/math] is defined for all x?
I guess this is fucking amateur hour.

I'd prove these trivialities to you but you're not paying me, so you can fuck off instead.

I dont care about your poor attempts at understanding analytical continuation, but please explain to me why you believe 1-1+1-1+1... = 1/2.

Even a first year university student can easily understand that this is completely and utterly wrong.

And as I said 1+2+3+4+5...= -1/12 BREAKS ANALYSIS.

Oh yeah? Prove it.

The sum

[math] \sum_{n=1}^\infty \left(\dfrac{1}{2 n-1} - \dfrac{1}{2 n} \right) [/math]

happens to be log(2).

As a side note, we can rewrite
[math] \dfrac{1}{c\, n-d} - \dfrac{1}{c n} [/math]
as
[math] \dfrac{d}{c\, n\, (c\,n-d)} [/math]
but that doesn't help much.

In any case
[math] \sum_{n=1}^\infty \dfrac{1}{2\, n} [/math]
or
[math] \sum_{n=1}^\infty \dfrac{1}{n} [/math]
or also
[math] \sum_{n=1}^\infty \dfrac{1}{c\, n} [/math]
for any constant c, is divergent. Therefore, we can't prove OPs claim in analysis. Even if both sides diverge.

As a side note pt. 2, even
[math] \sum_{n=1}^\infty \dfrac{prime(n)}{n} [/math]
where "prime(n)" is 1 when n is prime and otherwise sets the term to 0,
already diverges.

Shit, I'm literally maths major stoner dropout who failed baby's first real analysis and is currently on drugs, and even I know those are divergent series. What happened to you Veeky Forums?

...

They both diverge so technically they both equal infinity.

But the right one diverges a little faster.

>babby who passed intro to real anal thinks he knows anything
en.wikipedia.org/wiki/Ramanujan_summation

>poo math
lmao trashed u r a fukin nerd

The fact that -1/12 comes up as the sum of the naturals through 3 completely unrelated methods points at an interesting property of the naturals.

If you can't see it I feel bad for you.

Can you explain why this is wrong

Easy

[math] \sum_{n=1}^{\infty}n=-1/12 [/math]
[math] \sum_{n=1}^{\infty}1/n=-12 [/math]
[math] \sum_{n=1}^{\infty}1/(2n)+\sum_{n=1}^{\infty}1/(2n-1)=-12 [/math]

so by symmetry, we conclude,
[math] \sum_{n=1}^{\infty}1/(2n)=\sum_{n=1}^{\infty}1/(2n-1)=-6 [/math]

kek

makes sense, thanks

and then there's this idiot.

He's right, sort of. You can extend many functions on the complex plane defined by power series by giving an analytic function on a bigger set which agrees with your power series within the radius of convergence. While it's disingenuous to say that the sum is -1/12, there is a function which is sort of like the sum of 1/n^s which takes the value -1/12 at s=1.

you're an idiot, stop watching numberphile vids and instead, actually get through a fucking textbook

this is why we need math captchas boys and femboys

You have to finish the proof.
Def 1. By "plurivergent series" we mean one which is both divergent and convergent
Ex 1. Through comparison to the above, the harmonic series is plurivergent
Lemma 1. Plurivergent series do not exist.
The proof follows from the definitions of convergent and divergent series.
Corollary 2. The harmonic series does not exist
Thus the notion of vacuous truth supplies the main result:
Theorem 3. The series obtained from the harmonic series' even terms equals that obtained from its odd terms.
QED

the series [math]\sum_0^\infty \frac{1}{n}[/math] diverge.

with the same argument you could say
[math]\integral_{1}^{\infty}\frac{1}{x}dx = \infty[/math]
[math]\integral_{1}^{\infty}ln(x) dx = \infty[/math]
Thus
[math]\integral_{1}^{\infty}ln(x) dx = \integral_{1}^{\infty}\frac{1}{x}dx[/math]

Which is retarded and makes no sense at all.

>[math]\frac{1}{a + b} = \frac{1}{a}+\frac{1}{b}[/math]
neat

>Should be a piece of cake
sure, Grandpa

I'll bet.

logical questions should be easy too