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Why is this not convergent?

Anonymous Mon 15 Jul 02:15:05 2019 No.10807765 View Reply Original Report

Quoted By: >>10807772 >>10807805 >>10807920 >>10808346 >>10808349 >>10808485 >>10809275 >>10810254 >>10810567 >>10811943 >>10814360 >>10815615 >>10818942

> inb4 should have paid attention in class
>inb4 it's integral isn't convergent.

What is the proof that this is not convergent.

Anonymous

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Anonymous Mon 15 Jul 2019 02:19:00 No.10807772 Report

Quoted By: >>10807796 >>10807809

>>10807765
>inb4 it's integral isn't convergent.
>What is the proof that this is not convergent.
Well, then there's little we can do to help you.

Anonymous

Anonymous Mon 15 Jul 2019 02:33:16 No.10807796 Report

Quoted By:

>>10807772
Because there inst a number that represents that loop... Lets call it x and when that occurs it does converge.

Anonymous

Anonymous Mon 15 Jul 2019 02:38:57 No.10807805 Report

Quoted By: >>10807912 >>10809361 >>10810265 >>10810567 >>10812936

>>10807765
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6. . .
1/3 + 1/6 = 1/2
1/4 + 1/8 + 1/10 + 1/40 = 1/2
1 + 1/2 + 1/3 + 1/4... = 1 + 1/2 + (1/3 + 1/6) + (1/4 + 1/8 + 1/10 + 1/40) ....
= 1 + 1/2 + 1/2 + 1/2 +1/2 . ..
= 1 + 1 + 1 + 1...

Anonymous

Anonymous Mon 15 Jul 2019 02:39:53 No.10807809 Report

Quoted By: >>10807814 >>10807816 >>10807838

>>10807772

this explains nothing since it looks the same as any other convergent curve. And why can't this be the exception or the invalidation of the integral theory.

Anonymous

Anonymous Mon 15 Jul 2019 02:42:09 No.10807814 Report

Quoted By:

>>10807809
The picture shows that the sum is greater than the integral, so if the integral is infinite, the sum is infinite.

Anonymous

Anonymous Mon 15 Jul 2019 02:43:03 No.10807816 Report

Quoted By:

>>10807809
You should try and understand why the integral test works. It's pretty simple graphically, I'm sure there's a resource for it.

Anonymous

Anonymous Mon 15 Jul 2019 02:50:50 No.10807838 Report

Quoted By:

>>10807809
integral from 1 to infinity of 1/x = ln(inf)-ln(1) = infinity

the summation (as shown in the graph) is greater than the integral... thus it does not converge

Anonymous

Anonymous Mon 15 Jul 2019 03:27:02 No.10807911 Report

Quoted By: >>10809361 >>10810567 >>10812808

here is the proof

>defining the sum
s = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8.....
>collecting terms
s = 1 + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8)......
>changing some terms, creating an inequality
s > 1 + (1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8).....
>factoring out
s > 1 + 1*(1/2) + 2*(1/4) + 4*(1/8) + .....
>therefore
s > 1 + 1/2 + 1/2 + 1/2 + ......

which diverges and since s is greater, it also diverges

Anonymous

Anonymous Mon 15 Jul 2019 03:27:29 No.10807912 Report

Quoted By:

>>10807805
this makes sense.

Anonymous

Anonymous Mon 15 Jul 2019 03:33:54 No.10807920 Report

Quoted By: >>10810567

>>10807765
$1 \geq 1$
$1 + \frac{1}{2} \geq \frac{3}{2}$
$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \geq \frac{3}{2} + \frac{1}{4} + \frac{1}{4} \geq 2$
$1 + \frac{1}{2} + \cdots + \frac{1}{8} \geq 2 + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{5}{2}$

Continuing in this way, you can show by induction that the sum of the first 2^n terms is at least $1 + \frac{n}{2}$ , which gets arbitrarily large.

Anonymous

Anonymous Mon 15 Jul 2019 08:53:15 No.10808346 Report

Quoted By: >>10808357

>>10807765
This equals to the natural (base e) logarithm of n.

Anonymous

Anonymous Mon 15 Jul 2019 08:55:30 No.10808349 Report

Quoted By: >>10808784 >>10809216 >>10810567 >>10810575

>>10807765
$\displaystyle \begin{align*} \sum_{n=1}^{ \infty} \frac{1}{n} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\ &> 1 + \frac{1}{2} + \left (\frac{1}{4} + \frac{1}{4} \right ) + \left ( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right ) + \cdots \\ &= 1 + \lim_{n \rightarrow \infty} n \cdot \frac{1}{2} = \infty \end{align*} $

Anonymous

Anonymous Mon 15 Jul 2019 08:59:25 No.10808357 Report

Quoted By:

>>10808346
no it doesn't
https://en.wikipedia.org/wiki/E_(mathematical_constant)

Anonymous

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Anonymous Mon 15 Jul 2019 11:00:32 No.10808485 Report

Quoted By: >>10808495 >>10809811

>>10807765
Note that $\frac{1}{6}+\sum_{n=1}^\infty n = \frac{1}{6}-\frac{1}{12} = \frac{1}{12}$ . The inequality of harmonic and geometric means states $\frac{n}{1+\frac{1}{2}+\dots+\frac{1}{n}} \leq \frac{1+\dots +n}{n}$ for every integer $n \geq 1$ . It follows that $\frac{n^2}{1+\dots + n} \leq \sum_{k=1}^n \frac{1}{k}.$ But we have $\lim_{n\to\infty} \frac{n^2}{1+\dots + n} = \frac{1}{12}\lim_{n\to\infty}n^2$ so that the harmonic series diverges.

Anonymous

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Anonymous Mon 15 Jul 2019 11:06:04 No.10808495 Report

Quoted By:

>>10808485

Anonymous

Anonymous Mon 15 Jul 2019 14:32:27 No.10808784 Report

Quoted By:

>>10808349
this

Anonymous

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Anonymous Mon 15 Jul 2019 18:19:37 No.10809216 Report

Quoted By:

>>10808349

Anonymous

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Anonymous Mon 15 Jul 2019 18:48:58 No.10809275 Report

Quoted By:

>>10807765
Transform serie to integral form and found limit to 1 is infinite.
Use l'hopital rule

https://math.stackexchange.com/questions/451558/how-to-find-the-sum-of-this-series-1-frac12-frac13-frac14-do

Anonymous

Anonymous Mon 15 Jul 2019 19:37:13 No.10809361 Report

Quoted By: >>10809376 >>10809377 >>10809596 >>10809603 >>10809816 >>10813562 >>10814823 >>10817334

>>10807911
>>10807805
You can rearrange the harmonic series to "make it converge" to any other value. So this proof is incorrect.

Anonymous

Anonymous Mon 15 Jul 2019 19:44:56 No.10809376 Report

Quoted By: >>10809382 >>10809978

>>10809361
This isn't a rearrangment, the terms are in order. You cannot regroup any positive convergent series in the same way. Otherwise, for any positive number, you could find a number of terms of that series with sum greater than that number, i.e. the series would be divergent. It's not presented in the most rigorous (pedantic?) way, but it is correct.

Anonymous

Anonymous Mon 15 Jul 2019 19:45:28 No.10809377 Report

Quoted By: >>10809978 >>10810242

>>10809361
>You can rearrange the harmonic series to "make it converge" to any other value.
No you can't. Are you retarded?

Anonymous

Anonymous Mon 15 Jul 2019 19:46:46 No.10809382 Report

Quoted By:

>>10809376
this refers to the first proof only. the second proof can be made rigorous, but with more difficulty

Anonymous

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Anonymous Mon 15 Jul 2019 21:23:01 No.10809596 Report

Quoted By: >>10809978 >>10810242

>>10809361
You must be some special kind of retard. Quit maths and go make burgers. No tomatoes on mine

Anonymous

Anonymous Mon 15 Jul 2019 21:26:16 No.10809603 Report

Quoted By: >>10809978 >>10810242

>>10809361
Lol

Anonymous

Anonymous Mon 15 Jul 2019 23:08:37 No.10809811 Report

Quoted By: >>10810534

>>10808485
Wait I thought that the (-1/12) thing was a meme since the infinite summation of all real numbers is divergent ??!

Anonymous

Anonymous Mon 15 Jul 2019 23:09:40 No.10809816 Report

Quoted By: >>10809978

>>10809361
OP here, don't pin this shit on me.

Anonymous

Anonymous Mon 15 Jul 2019 23:35:55 No.10809868 Report

Quoted By: >>10815821

is there an intuitive way to rationalize why a series whos terms approach zero doesnt converge?
If it just is the way it is, then I understand there won't be an intuitive understanding, but I just can't see a reason why the harmonic diverges but the geometric converges. Is there a minimum rate of convergence or something where any larger series diverges?

Anonymous

Anonymous Tue 16 Jul 2019 00:17:52 No.10809978 Report

Quoted By: >>10810242

>>10809816
>>10809603
>>10809596
>>10809377
>>10809376
I think I had a bit of a brain fart lol. Riemanns theorem doesn't apply obviously.

Anonymous

Anonymous Tue 16 Jul 2019 02:36:11 No.10810242 Report

Quoted By:

>>10809978

It's the alternating harmonic series anon.

>>10809377
>>10809596
>>10809603

>didn't correct him correctly

Anonymous

Anonymous Tue 16 Jul 2019 02:42:05 No.10810254 Report

Quoted By: >>10810551

>>10807765
it's not divergent, it just converges to infinity-hat

Anonymous

Anonymous Tue 16 Jul 2019 02:49:43 No.10810265 Report

Quoted By: >>10810272 >>10810276

>>10807805
Wait this is illegal

Anonymous

Anonymous Tue 16 Jul 2019 02:52:34 No.10810272 Report

Quoted By:

>>10810265
Why

Anonymous

Anonymous Tue 16 Jul 2019 02:55:10 No.10810276 Report

Quoted By:

>>10810265
No it's legal.
It's a lower bound on a monotonically increasing sequence. The bound diverges so the sequence (of partial sums) diverges. Nothing illegal about that, it proves that it can not converge.

Anonymous

Anonymous Tue 16 Jul 2019 05:38:42 No.10810534 Report

Quoted By:

>>10809811
>real numbers
it's complex numbers

Anonymous

Anonymous Tue 16 Jul 2019 05:47:43 No.10810551 Report

Quoted By:

>>10810254
l,ao

Anonymous

Anonymous Tue 16 Jul 2019 06:00:15 No.10810567 Report

Quoted By: >>10812884 >>10815929 >>10817420 >>10817908

>>10807765
>all these bullshit answers
>all these calculus tests
>all these incomplete proofs
sigh. Real mathematicians forget everything the learned in calc 2 and just prove by M-N. You really expect me to memorize every little nuanced test in all of mathematics and know the proofs inside and out? fuck off. But what am I talking about, OP? This:
>>10807805
>>10807911
>>10807920
>>10808349
need to finish their proofs rigorously. That is all.

Anonymous

Anonymous Tue 16 Jul 2019 06:09:21 No.10810575 Report

Quoted By:

>>10808349
i would use k instead of n for the last equation for clarity but otherwise based

Anonymous

Anonymous Tue 16 Jul 2019 18:49:08 No.10811943 Report

Quoted By: >>10812040 >>10812759

>>10807765
I remember my Calculus 2 professor saying something along the lines of the sequence "not reaching 0 fast enough" and then gave the proof which I don't remember.

I wonder how much of that first bit was magical bs he sprinkled for brainlets like me.

Anonymous

Anonymous Tue 16 Jul 2019 19:45:49 No.10812040 Report

Quoted By: >>10815710

>>10811943
>not reaching 0 fast enough

That's honestly a pretty good answer.

Anonymous

Anonymous Wed 17 Jul 2019 01:06:43 No.10812759 Report

Quoted By: >>10815710

>>10811943
>he doesn't know about rates and orders of convergence
"not reaching 0 fast enough" is the only real world applicable answer.

Anonymous

Anonymous Wed 17 Jul 2019 01:27:17 No.10812808 Report

Quoted By: >>10812890 >>10813392

>>10807911
but if s=1+1/2+1/3...
s= 1.5.....
and 1+1/2+1/2+1/2... Has limit in infinity
so how s>1+1/2... If s will never reach 2 and operation on right can reach anything it want?
explain please, brainlet here

Anonymous

Anonymous Wed 17 Jul 2019 01:43:04 No.10812844 Report

Quoted By: >>10812895

This series has limit in infinity, because for any n E 1/n >= every k E 1/k where k =< n? Am i understanding this correctly?
brainlet here i want to kearn

Anonymous

Anonymous Wed 17 Jul 2019 02:04:35 No.10812884 Report

Quoted By: >>10813587

>>10810567
those are rigorous. literally just showing that a subsequence of the partial sums is bounded below by a divergent sequence. subsequence diverges implies the sequence diverges. stop being a moron. you aren't fit for math if you forget basic intuitive shit like this.

Anonymous

Anonymous Wed 17 Jul 2019 02:06:24 No.10812890 Report

Quoted By:

>>10812808
you're just wrong. 1 + 1/2 + 1/3 + 1/4 is already greater than 2.

Anonymous

Anonymous Wed 17 Jul 2019 02:08:47 No.10812895 Report

Quoted By:

>>10812844
no one knows what you're writing. are those supposed to be existentials?
the limit is infinity because for any R > 0 there exists an N such that for all n bigger than N the sum 1 + 1/2 + 1/3 + ... + 1/n is greater than R. In other words, the sum gets larger than any positive real number as long as you go far enough out.

Anonymous

Anonymous Wed 17 Jul 2019 02:28:19 No.10812936 Report

Quoted By: >>10814190 >>10817141

>>10807805
1-(1/2)+(1/3)-(1/4)+...
= 1+(1/3)+(1/4)+... = Infinity
= -(1/2)-(1/4)-... = -infinity

So infinity = -infinity

Sweet! Thanks Mom!

Anonymous

Anonymous Wed 17 Jul 2019 04:32:11 No.10813392 Report

Quoted By:

>>10812808
>s=1+1/2+1/3...
>s= 1.5.
wew lad

Anonymous

Anonymous Wed 17 Jul 2019 05:27:04 No.10813562 Report

Quoted By:

>>10809361
No you cant
all the terms are positive, so, if you could rearrange the terms to get them to converge, then it would be absolutely convergent
and if it is absolutely convergent, then the series would only converge to a single limit no matter how you rearrange it.
So, the fact that you can find one rearrangement of the terms such that their sum diverges to infinity, proves that all rearrangements will diverge to infinity, including the original series.

Anonymous

Anonymous Wed 17 Jul 2019 05:39:55 No.10813587 Report

Quoted By: >>10813918

>>10812884
>showing that a subsequence of the partial sums is bounded below by a divergent sequence
no rigorous proof of this was shown.
>hint
ellipses are not allowed in your argument.

Anonymous

Anonymous Wed 17 Jul 2019 09:10:38 No.10813918 Report

Quoted By: >>10815885

>>10813587
>pulling rules out of my ass
the post

Anonymous

Anonymous Wed 17 Jul 2019 11:42:40 No.10814190 Report

Quoted By: >>10814496

>>10812936
The sum of 1/2n (your examples) is convergent

Anonymous

Anonymous Wed 17 Jul 2019 13:41:01 No.10814360 Report

Quoted By:

>>10807765
It os not convergent, because if you tell me a number, i can tell you another that is the index of the sum that exceeds it.

Anonymous

Anonymous Wed 17 Jul 2019 14:51:50 No.10814496 Report

Quoted By: >>10814502

>>10814190
The sum of 1/2n as n goes to infinity is not convergent.

Anonymous

Anonymous Wed 17 Jul 2019 14:55:37 No.10814502 Report

Quoted By: >>10814609

>>10814496
1+.5+.25+.125+...
is convergent

Anonymous

Anonymous Wed 17 Jul 2019 15:44:07 No.10814609 Report

Quoted By:

>>10814502
That's 1/(2)^n, not 1/2n.

Sage

Sage Wed 17 Jul 2019 17:21:51 No.10814823 Report

Quoted By:

>>10809361
That would be true if the terms were alternating in sign. They're all positive so it's clearly divergent (and hence not approaching any real value as it goes to infinity), hence it's a valid proof, granted a bit lacking, but the average mathematician can fill in the gaps.

Anonymous

Anonymous Wed 17 Jul 2019 23:09:42 No.10815615 Report

Quoted By: >>10815706

>>10807765
If you can graph functions this shouldn't be hard

Anonymous

Anonymous Wed 17 Jul 2019 23:53:49 No.10815706 Report

Quoted By:

>>10815615
How is it apparent in a graph?

Anonymous

Anonymous Wed 17 Jul 2019 23:57:31 No.10815710 Report

Quoted By: >>10815885 >>10815944 >>10815975 >>10817848

>>10812040
>>10812759
it's a completely tautological answer. how fast is "fast enough"?

Anonymous

Anonymous Thu 18 Jul 2019 01:08:25 No.10815821 Report

Quoted By:

>>10809868
Isn't it intuitive enough to just say, unlike geometric, the terms just don't decrease fast enough?

Anonymous

Anonymous Thu 18 Jul 2019 01:42:22 No.10815885 Report

Quoted By: >>10816292

>>10813918
Here's a rigorous proof that you are a brainlet and a fag, following your rules:
>heh, this guy
>...
>posting on a japanese gay tentacle hentai site...
>not very smart because the contents of his posts...
>QED
>>10815710
I don't know, because I am not an expert at that kind of math, and perhaps "fast enough" may not even follow formula for a metric depending on one variable, but it definitely is a legitimate answer. I suggest going back to the prof who said it and asking him to go more in depth.

Anonymous

Anonymous Thu 18 Jul 2019 01:58:29 No.10815929 Report

Quoted By:

>>10810567
rigorous? here ya go

http://us.metamath.org/mpegif/harmonic.html

Anonymous

Anonymous Thu 18 Jul 2019 02:02:33 No.10815944 Report

Quoted By:

>>10815710
1/fractal?

Anonymous

Anonymous Thu 18 Jul 2019 02:16:16 No.10815975 Report

Quoted By:

>>10815710
> how fast is "fast enough"

1/n^p where p is larger than 1

Anonymous

Anonymous Thu 18 Jul 2019 05:35:35 No.10816292 Report

Quoted By:

>>10815885
seething

Anonymous

Anonymous Thu 18 Jul 2019 17:00:53 No.10817141 Report

Quoted By:

>>10812936
>What is convergence without absolute convergence?

Anonymous

Anonymous Thu 18 Jul 2019 18:46:01 No.10817334 Report

Quoted By:

>>10809361
Not a rearrangement, (no abuse of commutativity) just associativity. To rigorize, assume the series converges to S. Let e>0 and take a partial sum value P s.t. |P-S|<e, with the same inequality for sums with more terms than P. Show that there exists one sum using associative manipulation above breaking the given inequality.

Anonymous

Anonymous Thu 18 Jul 2019 19:36:49 No.10817420 Report

Quoted By:

>>10810567
>need to finish their proof rigorously
Just imagine that they added "the remaining is trivial and left as an exercise to the reader"

Anonymous

Anonymous Thu 18 Jul 2019 22:52:02 No.10817848 Report

Quoted By:

>>10815710
it's a very fuzzy boundary. you can get really really close on either side, but i don't think there is such an order of growth such that series above it diverge and series below it converge.
one can have, for instance, the terms
1/(n*log(n)) or the terms 1/(n*log(n)*log(log(n))) or series whose terms contain further iterates of log, and your sequence will always diverge. similarly, you can approach incredibly slow convergence from below.
i don't know if there is theory regarding this, but i'd love to learn more about it.

Anonymous

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Anonymous Thu 18 Jul 2019 23:13:51 No.10817908 Report

Quoted By:

>>10810567
I'll give you a rigorous proof if you'd like.
Denote the partial sums of the harmonic series by $H_n$ . We aim to show divergence, in fact divergence to $+\infty$ . We pass to the subsequence $A_k = H_{2^k}$ . Note that $A_0 = 1$ . Then for $k > 0$ we find $A_k = \sum_{j=1}^{2^k} \frac{1}{j} = \sum_{j=1}^{2^{k-1}}\frac{1}{j} + \sum_{j=2^{k-1}+1}^{2^{k}}\frac{1}{j} \geq A_{k-1} + \sum_{j=2^{k-1}+1}^{2^{k}}\frac{1}{2^k} = A_{k - 1} + 2^{k-1}\frac{1}{2^k} = A_{k-1} + \frac{1}{2}$ . Thus $A_k \geq A_{k - 1} + \frac{1}{2}$ . Summing the inequalities from $k = 1$ to $N$ , we find that $A_{N} \geq 1 + \frac{N}{2}$ . Thus, for each $R > 0$ , pick $N > 2R$ and then $A_N \geq 1 + \frac{N}{2} > R$ . This shows that $A_k$ diverges to $+\infty$ .
Of course, it is a basic fact that divergence of a subsequence implies divergence of the sequence $H_n$ . Furthermore, $H_n$ is monotone, and monotone divergent series must diverge to $+\infty$ . This completes the proof.

Anonymous

Anonymous Thu 18 Jul 2019 23:20:28 No.10817933 Report

Quoted By:

The supremum of partial sums does not exist (1, 3/2, 11/6,...)

Anonymous

Anonymous Fri 19 Jul 2019 09:51:52 No.10818942 Report

Quoted By:

>>10807765
proved by a 13th century Monk by comparing partial sum with 0.5

that Monk has very impressive algebraic talent, that Monk deserve a place in mathematics

Anonymous

Anonymous Fri 19 Jul 2019 09:54:26 No.10818945 Report

Quoted By:

1/3+1/4 > 1/4+1/4=1/2
1/5+1/6+1/7+1/8 > 1/8+1/8+1/8+1/8=1/2
...
1/(2^n+1) + ... + 1/2^(n+1) > 1/2^(n+1)+...+1/2^(n+1) = 1/2
...

sum all the numbers we have 1/3+1/4+...+1/infinity > 1/2 + 1/2 + 1/2 + ......
so the sum diverges

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