Another challenging series for 4chan

[Anonymous No.10254173]

Does it converge or diverge?

[Anonymous No.10254180]

converge

[Anonymous No.10254191]

https://en.wikipedia.org/wiki/Dirichlet_series#Analytic_properties_of_Dirichlet_series
Knock yourselves out.

[Anonymous No.10254197]

>>10254173
Converge b/c 1/n^p converges when p>1, 1+|lim n|>1

[Anonymous No.10254219]

>>10254197
I understand your point, but you are not considering that sin(n) can approximate infinitely well 0, as n goes to infinity. and this fact most likely makes the series diverge, as it seems from computing the partial sums.

[Anonymous No.10254229]

>>10254219
But that's |sin(n)| and there is no way you can say that's 0 in the limit.

[Anonymous No.10254232]

>>10254229
I'm not saying that, and I see what you're saying but I actually think that it doesn't work in this case. I don't think it converges because there are terms in which sin(n) gets close to zero as much as you like yielding some terms that behave like 1/n which make the sum diverge

[Anonymous No.10254233]

>>10254229
I'll say it better.
As you move p downwards closer and closer to 1, the sum expands without limit. You can bound the sum with any of these up until some N, but then it breaks through, so your proof doesn't count.

[Anonymous No.10254248]

we had this kind of question many times
use basic trigonometry to see that the convergence is equivalent to a statement about how well pi can be approximated by rational numbers
then read about continued fractions and rational approximations to irrational numbers, and the problem is solved

[Anonymous No.10254249]

>>10254232
But all of the terms would be bigger than zero nonetheless since sin(n) cannot ever be 0 for the values of n you're summing over. Even though some of the terms approximate zero, none of them are and I'm pretty sure that's what matters. Good luck calculating what it converges to though.

[Anonymous No.10254253]

>>10254197
>>10254232
friendly remainder that f(n) > 0 and f(n) >= k, where k > 0, are two very different things in analysis.

[Anonymous No.10254258]

>>10254173
converge

50 50 chance

[Anonymous No.10254259]

>>10254249
but it is not a fixed value. if instead of having 1 + abs(sinn) you had 1 + 1/n, you'd definitely say that it diverges right? because even though 1/n is never zero as you take the limit it tends to zero. what I'm saying is that abs(sin(n)) tends to zero as much as it tends to any other number between 0 and 1 when you go to infinity, although in this case I am interested in 0 because it makes the series diverge

[Anonymous No.10254261]

so many retards in this thread not knowing how to apply simple mean value theorem to solve this lmao

[Anonymous No.10254265]

>>10254253
exactly. that's why it diverges, imo.

[Anonymous No.10254266]

>>10254265
it converges but it's not an absolute convergence brainlet

[Anonymous No.10254267]

>>10254266
> it's not absolute convergence
the terms are always positive.

[Anonymous No.10254269]

>>10254249
>bounding a sequence with a diverging sequence of bounds
For fuck's sakes man, is your brain made out of peanut butter?
>>10254261
Mildly curious.

[Anonymous No.10254272]

>>10254269
replace the sin function with something "worse" to prove that it still converges

[Anonymous No.10254273]

>>10254272
elaborate

[Anonymous No.10254277]

>>10254273
what's the worst value that sin(n) can assume in terms of divergence?

[Anonymous No.10254279]

>>10254277
0

[Anonymous No.10254283]

>>10254279
and the best one?

[Anonymous No.10254285]

>>10254283
1

[Anonymous No.10254286]

>>10254283
why do I feel like you are getting to nowhere? it diverges.

[Anonymous No.10254289]

>>10254173
Sorry, xir, this is not 4chan. This is /r/4channel

[Anonymous No.10254293]

>>10254258
>50 50 chance
It’s 2/3

[Anonymous No.10254295]

>>10254286
>>10254285
>>10254279
now my dear retards, what is the lowest value instead of sin(n) where the sum still converges?

[Anonymous No.10254298]

>>10254173
I think it converges. It’s “smaller” than the harmonic series.

[Anonymous No.10254300]

>>10254295
anything >k for any k>0. so what you just said is utterly useless since we have 0 that makes the sum diverge.

[Anonymous No.10254301]

This sum converges iff the sum of 2^(-n |sin n|) converges, which might be easier to check. I haven't tried yet.

[Anonymous No.10254303]

>>10254298
1/(nlog n) is smaller than the harmonic series and yet diverges.
I can find even smaller.

[Anonymous No.10254305]

>>10254300
does sin(n) always get 0? because unless it does it converges you absolute mongoloid

[Anonymous No.10254311]

>>10254295
The minima of sin(n) is zero.
>>10254305
>bounding
>a
>sequence
>with
>a
>divergent
>sequence
It absolutely has to be retardation.

[Anonymous No.10254312]

>>10254305
oh so it needs to be always zero? where the fuck did you learn math? retard.

[Anonymous No.10254315]

>>10254305
You're a fucking idiot, you know that? Please stop shitting up this thread. Just because a series is smaller than a divergent series doesn't tell you shit. At the moment there is no information as to whether the series is smaller than any convergent series, because it has terms which get larger than any convergent 1/n^p series for infinitely many n. Do you just not understand high school analysis?

[Anonymous No.10254316]

>>10254312
yes it always needs to be zero in a periodic function to diverge. where did YOU learn math

[Anonymous No.10254318]

>>10254266
Are you retarded

[Anonymous No.10254320]

>>10254301
So it comes down to proving that n sin n can get arbitrarily close to 0 for large enough n. Then one can take a subsequence which does not go to 0. Fuck, this is weird. Is that true? Seems like it should be.

[Anonymous No.10254322]

>>10254320
No, it can still flip around arbitrarily close to 0 and converge.

[Anonymous No.10254325]

>>10254316
It won't always be zero but will approximate 0 infinitely many times and will make the series diverge.

[Anonymous No.10254326]

>>10254322
You didn't read my previous post, did you? I used the dyadic trick so that the series 2^(-n*|sin n|) converges iff op's series converges. If n*sin n gets arbitrarily close to 0, the sequence does not limit to 0. At the moment, the goal is to show that 2^(-n * |sin n|) does not have a limit of 0.

[Anonymous No.10254327]

>>10254325
an approximation to zero does not make it converge anon. jesus this is first semester math, anything, even 10^-21 above 0 makes a harmonic series converge

[Anonymous No.10254329]

>>10254327
diverge* instead of the first converge

[Anonymous No.10254331]

>>10254327
I know, why don't you read the previous posts? it has already been discussed. the thing is, sin(n)>0, but it's not true that always sin(n)>k where k>0, because you can always get arbitrary closer, just like 1/n>0 but you can get arbitrarly close to zero. You can't treat like a constant like in your example, you have to see it as a limit. that's what it actually is.

[Anonymous No.10254336]

>>10254325
Literally incorrect.
For example, 1/n^2 is in fact a subsequence of 1/n, but it still converges. Just because the sum in the OP "meets" 1/n enough times doesn't mean it'll also diverge. That's wrong.
>>10254326
Give me proof of your previous statement.
Or at least a link.

[Anonymous No.10254337]

>>10254327
>>10254329
Not the person you're arguing with, but you're both idiots for thinking it definitely converges / diverges since on one hand you can't bound it by one of your 1 + 1/10^21 sequences but on the other the terms approaching zero might spread out very thin, in which case it still could converge.
My dyadic trick argument might resolve this.

[Anonymous No.10254344]

>>10254336
Aw fuck, that trick only works for monotone decreasing sequences. Damn.
Ignore my bullshit.
If the sequence were decreasing, we would have the sum a_n converges iff the sum 2^n a_(2^n) converges by a simple two-way comparison.

[Anonymous No.10254345]

>>10254344
It's okay, you were still one of the smarter people in the thread.

[Anonymous No.10254379]

it converges to something smaller than or equal to pi+1

/thread

[Anonymous No.10254386]

>>10254293
Brainlet

[Anonymous No.10254429]

are the people in here unironically retarded? there is no natural number where sin(x) will ever reach 0. thus the series converges because the exponent is always greater than 1.
this can be easily proven with a cauchy condensation test, just assume it says alpha instead of 1+sinwhatever and then prove that it converges if alpha>1 which is always the case.
>>10254249
is the only non brainlet.

[Anonymous No.10254440]

>>10254429
gtfo if you don't even bother attempting to understand what has been said over and over in this thread.

[Anonymous No.10254448]

>>10254440
mad brainlet detected
I bet you don't even know what a condensation test is and why it matters here

[Anonymous No.10254456]

>>10254301
shouldn't it be 2^(-nsin(2^n))? maybe it doesn't make a difference, but still

[Anonymous No.10254461]

>>10254448
I know very well what it is, why don't you just fucking read the previous posts and understand why your argument doesn't hold?

[Anonymous No.10254462]

>>10254429
Given an [math] \alpha > 1 [/math] there are infinitely many [math] n [/math] such that [math] 1 + |\sin{n}| < \alpha [/math] so you cannot just compare it to [math] 1/n^\alpha [/math].

[Anonymous No.10254476]

>>10254456
It should be, but it doesn't matter anyway. See my post above. Needs to be monotone.
>>10254429
You're fucking unsalvageable. If you need a concrete example, take pn to be pn = (log log n)/(log n) + 1. pn > 1 for all large enough n, but 1/n^(pn) = 1/(n*n^((log log n)/(log n))) = 1/(n*log n) which has a divergent series.

[Anonymous No.10254477]

>>10254429
Imagine posting this three (3) posts after someone noticed that the sequence isn't monotone decreasing.

[Anonymous No.10254488]

>>10254440
>>10254462
>>10254476
>>10254477

[Anonymous No.10254493]

>>10254488
What? No! alpha is a constant there! Totally different!
The condensation test only holds for monotone decreasing sequences, and if you replace alpha by alpha(n), the sequence is no longer necessarily monotone decreasing (not does it necessarily converge if it were).

[Anonymous No.10254502]

>>10254191
>be me
>post the solution method
>ayy lmao no one does anything with it
>do some shitposts correcting people for the sake of it
Ah, whatever.
Anyhow, [math]a_n=n^{-|sin n|}[/math].
Larger than the harmonic series, so its sum diverges.
Using a nice and pretty formula in the wikipedia page, [math]lim~sup~ln(\sum_{n=1}^{N} a_n)/ln N[/math] gives the abscissa of convergence. Since the sum grows faster than the harmonic, we have a value bigger than 1, and it diverges at 1.

[Anonymous No.10254511]

>>10254502
>larger than the harmonic series
literally impossible, learn to read brainlet

[Anonymous No.10254516]

>>10254493
The sequence needs to be non-increasing and as >>10254493 said alpha is constant. You are a joke.

[Anonymous No.10254520]

>>10254516
>muh specific rules don't apply thus its false
what part about FOR ALL do you not understand, brainlet. The p-series converges for ALL p>1.
Can you name one instance in the sum where p=1?

Thought so.

[Anonymous No.10254527]

>>10254502
My bad. Each element of the sum is smaller than 1, so we have a ratio below one, and the sum converges.
>>10254520
>we can bound it with an increasing sequence of bounds, thus it's bounded
This is your psychological on heroin.

[Anonymous No.10254530]

>>10254520
>muh specific rules don't apply thus I can't use a theorem
That's typically how math works yes.

[Anonymous No.10254534]

>>10254520
listen retard, since there is no way you can understand this, compute a bunch of partial sums and see for yourself how the sum behaves like it diverges.

[Anonymous No.10254535]

>>10254511
t. didn't read his post
>>10254502
i didn't bother reading through that wiki page before because it's written by a fucking moron, but yeah, it seems like you're right. well done.
i'm not 100% convinced about the limit though. i think you've bounded top series below by log N, so you're claiming log log N / log N has a limsup greater than 1 which is false. you need to bound the top series below by N (which i don't think you'll manage).

[Anonymous No.10254537]

>>10254527
you have to understand the principles behind differentials before clinging to your retarded rules, brainlet.
>>10254520
is also correct, although you technically "can't do it" with a integral or bounding test, you will still arrive to the same result each and every time.

[Anonymous No.10254539]

>>10254527
ah, you answered my question >>10254535 so it seems like you're right.
can we be sure that the limit is not 1 though? does the formula guarantee convergence at the abscissa?

[Anonymous No.10254540]

>>10254535
Yeah, I noticed it later, and corrected it here >>10254527
Each a_n is smaller than 1, so as you sum them until N, the sum is smaller than N.

[Anonymous No.10254545]

>>10254502
where did you learn this stuff? I still have never heard of the abscissa of convergence. just curious.

[Anonymous No.10254548]

>>10254520
>>10254537
FUCKING WRONG. LOOK AT MY FUCKING SERIES YOU MORONS ARE CONVENIENTLY IGNORING.
>>10254476
THIS SATISFIES YOUR FUCKING BULLSHIT CONDITION, AND IT DIVERGES.
NEVER FUCKING TOUCH ANALYSIS AGAIN, YOU IDIOTS!

[Anonymous No.10254549]

>>10254539
>the limit is not 1
About 2 in three values are below 1/2, so probably not.
I haven't proof tho. Messing with sines sucks.

[Anonymous No.10254552]

>>10254545
Looking at the Riemann Hypothesis page, saw the Grand Riemann Hypothesis, looked up L-series and Dirichlet series for the fun of it.

[Anonymous No.10254553]

>>10254549
you're probably right, although just to be sure we should probably check. the wiki page doesn't tell us anything about convergence AT the abscissa.

[Anonymous No.10254565]

op pls absolve us from this bullshittery and tell us the solution

[Anonymous No.10254584]

>>10254549
wait, you can't seriously intend to make me believe that this sum converges with 1/2 + |sin n| instead of 1 + |sin n|
that fucking abscissa is not 0, though it's probably below 1

[Anonymous No.10254591]

>>10254549
>>10254584
https://www.desmos.com/calculator/nkqsd7cswe
clusters around 0.66, so it's almost definitely below 1
nicely done

[Anonymous No.10254594]

>>10254584
2 in three values of n^(-|sen n|).
The logarithm is definitely positive tho.

[Anonymous No.10254601]

>>10254594
Yeah, 2/3 is pretty much the answer >>10254591 (though not exactly it would seem - i don't know, this doesn't converge very quickly)
it would be nice to bound it for real

[Anonymous No.10254602]

>>10254565
I am op and have been participating in the discussion this whole time. I don't have any proof but definitely feel like it diverges for the reasons that have been discussed in the thread multiple times. I also have computed a partial sum to n=1000, pic related. with n=10000 it yields 2.8something. so yeah it seems like it diverges.

[Anonymous No.10254605]

>>10254591
When N = 10000000, it's about 0.8 though

[Anonymous No.10254609]

>>10254601
Oh fuck
https://www.desmos.com/calculator/7khpda8yda
ITS FUCKING GROWING

[Anonymous No.10254612]

>>10254605
that's what i just noticed >>10254609
actually quite worried this thing goes to 1, though that's extremely surprising.

[Anonymous No.10254613]

>>10254612
This >>10254602 shows it outgrows 1.
It's probably divergent.

[Anonymous No.10254616]

>>10254613
that's not the same thing we're looking at, we're looking at the abscissa of convergence
just because the sum grows past 1 doesn't mean it diverges for fucks sake

[Anonymous No.10254727]

>>10254488
this is unironically correct, however it is missing a crucial part.
we know that sin(n) will never be 0, however since the exponent is not a constant we cant use cauchy or integral or anything like that. So we'll just assume instead that the smallest possible sin(natural number). this obviously converges, so the series converges either way.

[Anonymous No.10254729]

>>10254727
*smallest possible sin is the constant

[Anonymous No.10254731]

>>10254727
>the smallest possible sin(natural number)
and what's this?
inf{sin(n) : n natural} = 0

[Anonymous No.10254735]

>>10254731
whoops, i mean |sin n|
i can provide a simple proof

[Anonymous No.10254736]

>>10254735
sin(x)=0 can never be a natural number since it always has pi as a component. but sure, show me your proofs

[Anonymous No.10254748]

no correct approaches to the solution so far.
it diverges

[Anonymous No.10254750]

>>10254748
I'm too brainlet to understand this

[Anonymous No.10254753]

>>10254736
he's not saying |sin(n)| = 0, he's saying that the infimum of the set of all solutions is 0.

[Anonymous No.10254755]

>>10254736
that's not what an infimum is, there is no minimum.
here is a proof:
let epsilon > 0. since sin is uniformly continuous, there exists the relevant delta > 0.
it is a well known result ( https://math.stackexchange.com/questions/24806/multiples-of-a-given-irrational-number-can-be-arbitrarily-close-to-a-natural-num ) that we can find natural m, n s.t. |m - npi| < delta. Then |sin(m) - sin(npi)| = |sin(m)| < epsilon.
Thus |sin(m)| can be arbitrarily small for large enough m.
Of course, m does not equal npi for any m, n, so sin(m) is never 0. Thus sin(m) never attains a minimum.

>>10254748
this is fucking retarded, what an awful question

[Anonymous No.10254756]

>>10254727
>all this work to say what I said >>10254298

[Anonymous No.10254759]

>>10254755
>this is fucking retarded, what an awful question
it is the official solution tho.

[Anonymous No.10254769]

>>10254759
sorry, i saw 7n and flipped my shit
i didn't realize it was just cause 7 is above 2pi
it's actually quite beautiful
>>10254756
but it diverges though, look at the official solution
fucking p-series idiots btfo

[Anonymous No.10254773]

>>10254748
now that i've actually read through this and understood it, it's fucking wonderful
i love it, thanks anon

[Anonymous No.10254779]

>>10254748
Can someone explain this for us brainlets?

[Anonymous No.10254781]

because [math] n \in \mathbb{R} [/math] and there will be no negative values to the sign functions, there will always be something strictly greater than one on the exponent as in [math] 1 + |\sin{(n)}| > 1 [/math] thus obviously converging.

this can be further proved by applying pretty much any convergence test that isn't a comparison test. regardless, I think my intuition is correct.

[Anonymous No.10254784]

>>10254781
incorrect, as shown already by >>10254748
stop larping, calc 1 idiots

[Anonymous No.10254785]

>>10254781
fuck, meant to say [math] n \in mathbb{N} [/math]

[Anonymous No.10254795]

>>10254748
>this grows without bound
Doesn't |A_1| tend to infinity, tho?

[Anonymous No.10254805]

>>10254779
sure, here's an idea of what's going on
we look at all the naturals below 2^n and just consider those for which the sin is small. we then show by a pidgeonhole argument that at least 2^n / (7n) of these exist (to see this, think about spinning a circle a set angle 2^n times, it should get close to it's original position once in a while, and as the problem proves, at least 1/(7n) of the time for rotation of n radians). this number of points where the sin is small is called |A_n|. from there we split the sum into a bunch of parts, from 2^(n-1) to 2^n, and we know each part is bigger than (|A_n| - |A_(n-1)|) / 2^(n+1) by bounding. This sum telescopes (partially) and we end up with some garbage on the front and a |A_n|/2^(n+2) still in the sum, which is more than (2^n / 7n) / 2^(n+2), which is 1/28 times the harmonic series and thus diverges.

[Anonymous No.10254811]

>>10254795
how? |A_1| is a fucking constant. it's just the set of integers between 0 and 1 for which sin(n) < 1, which is the set {0, 1}. so |A_1| = 3.

[Anonymous No.10254817]

>>10254811
*|A_1| = 2 i mean

[Anonymous No.10254830]

>>10254805
Cool, thanks anon.

[Anonymous No.10254916]

I thought p-series always converged as long as the exponent was greater than 1.
having made some rudimentary code to test for a shit ton of iterations this series very well seems to diverge, but what do I know, im just an engineer.

my code:

package series;

import java.util.Scanner;

public class main {

public static void main(String[] args) {

String cont = "y";
Scanner sc = new Scanner(System.in);

System.out.println("Enter number of iterations");
long iter = sc.nextLong();

double result = 0;
for (int n = 1; n <= iter; n++) {
double k = 1 + abs(Math.sin(n));
result += 1/(Math.pow(n, k));
}

System.out.println("the partial sum for " + iter + " interations is: " +result);
}

private static double abs(double a) {
if (a >= 0)
return a;
else
return -a;
}
}

[Tuba Jam No.10254984]

>>10254552
based touhou anon, how are you so smart? When did you first get into mature math?

[Anonymous No.10254995]

>>10254984
That's definitely not Tohou anon

[Anonymous No.10255005]

>>10254984
Yukariposter is the super smart Touhou anon, I'm the "middling at best" Touhou anon who's trying to use voodoo to steal the former's intelligence.

[Anonymous No.10255010]

>>10254916
this is not a p-series
p-series do converge if the exponent is greater than 1, but p-series have a constant exponent
this is a dirichlet series, which is similar but more general

[Anonymous No.10255014]

>>10255005
yukariposter-sama is my role model!
when i grow up, i wanna be just like yukariposter-sama!

[Anonymous No.10255020]

>>10255010
>more general
Y'all niggas say the weirdest things.
>>10255014
No, not to that extent, it's just a joke.

[Tuba Jam No.10255027]

>>10255020
I don't think >>10255014 was mocking you.
I'm still curious about how you gained your mathematical maturity. Not anyone can self-learn stuff like Dirichlet series like that. Are you an undergrad? Are you at an Ivy?

[Anonymous No.10255039]

>>10255027
This was my freshman year.
I just knew that stuff by coincidence. Besides, the solution was apparently wrong (because apparently, that series actually converges to 1 I mean what the fuck.)
Everyone who wants to eventually hits mathematical maturity. Literally anyone could learn stuff like Dirichlet series.

[Anonymous No.10255053]

>>10255027
stop sucking that guy's dick, if you look into RH for more than 10 minutes you run into dirichlet series. lord almighty.
>>10255020
what's the issue with "more general?" p-series are quite literally dirichlet series with a_n = 1 (so the zeta function) and a real variable.

[Anonymous No.10255065]

>>10255053
Based and correct.
Also, my bad, I'd misremembered the p-series definition.

[Anonymous No.10255246]

>>10254173
Impossible to prove

[Anonymous No.10255248]

>>10254916
DESIGNATED
SHITTING
JAVA

[Anonymous No.10255253]

>>10254748
Where is this solution from?

[Anonymous No.10255264]

>>10255248
What?

[Anonymous No.10255308]

So I plotted the sum vs n for this problem and 1/n for reference, for n up to 1000000. Looks like it's growing but at a snails pace because it only really grows when |sin(n)| approximates 0, but since it does so infinitely many times it doesn't converge. Counterintuitive because |sin(n)| is technically always bigger than 0 but what is intuitive in maths nowadays.

[Anonymous No.10255323]

>>10255308
Do a few shanks transformations to accelerate the convergence.

[Anonymous No.10255328]

>>10255308
sorry, how is it counterintuitive?
or you're just a brainlet when it comes to analysis, huh?
my example here >>10254476 is a completely intuitive example of the exponents all being greater than 1, but the series diverging. you moron.

[Anonymous No.10255333]

>>10255328
Idk, haven't done Analysis for 4 years.

[Anonymous No.10255652]

Converges. |sin(n)| > 0 for all n in the positive integers as there is no integer solution for |sin(n)|=0 and the series 1/(n^p) converge for all p>1

[Anonymous No.10255656]

>>10255652
read the thread

[El Arcón No.10255677]

>>10254322
>it can still flip around arbitrarily close to 0 and converge.

[El Arcón No.10255681]

>>10254326
>You didn't read my previous post, did you?
Tempus Edax Rerum
http://www.vixra.org/abs/1209.0010

[El Arcón No.10255691]

>>10254488
i know right?

[Anonymous No.10255696]

>>10255681
Based schizo responding to my post with something completely incidental.
>>10255652
How did so many complete morons find their way to /sci/? I'd love to see you try to prove the p-series convergence condition replacing p by a function. It'll be funny to watch.

[El Arcón No.10255720]

>>10255039
>Everyone who wants to eventually hits mathematical maturity.
I agree

[Anonymous No.10255773]

>>10254173
A BUNCH OF FUCKING SHIT

[Anonymous No.10255837]

>>10254173
converges. exponent is always > 1 since n is an integer > 0 and therefore 0 is out of range of |sin|.

[Anonymous No.10255858]

>>10255837
read
>>10255656

[Anonymous No.10255962]

it clearly converges to some value(s) <= inf

[Anonymous No.10256433]

>>10255308
Thanks for the computation, looks like the predictions made early in the thread hold up. Btw I am op and pretty proud that this thread generated such response.

[Anonymous No.10256486]

>>10256433
on the matter has a correct formal proof been eventually found? maybe >>10254502 ?

[Anonymous No.10256508]

>>10256486
See >>10254748
And every other retard that recently popped up that says it converges should have a look.

[Anonymous No.10256548]

>>10256508
>>10254502's proof isn't valid?

[Anonymous No.10256566]

Smart people assert that it converges because it's obvious while useless brainlets argue about mathematical proofs.

[Anonymous No.10256573]

>>10256566
t. Retard that applies theorems blindly without really understanding the core of analysis.

[Anonymous No.10256755]

>>10254429
friendly reminder that f(n) > 0 and f(n) >= k, where k > 0, are two very different things in analysis.

[Anonymous No.10257388]

>>10256548
It's all valid until the end, but we can't conclude convergence. The limit must converge to 1. We cannot conclude convergence or divergence at the abscissa of convergence.
We all thought that thing converged to something less than 1, but it has to converge to 1 since it's below 1 but the series has been proven to diverge. Therefore the abscissa is 1 (and it diverges at the abscissa).