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You should be able to solve this problem (standard for highschool students in countries where people poo in the loo)
>>9765300diverges because oddsin term is irrelevant
>Reminder: /sci/ is for discussing topics pertaining to science and mathematics, not for helping you with your homework or helping you figure out your career path.>If you want help with your homework, go to /wsr/ - Worksafe Requests.
Please stop being a faggot on this board, just look up calc 2 on the interwebs.saged
>>9765300It has an over-harmonic majorant, so it converges.
>>9765307>faggotWhy the homophobia?
http://mathworld.wolfram.com/FlintHillsSeries.html>It is not known if this series converges
>>9765307>just look up calc 2>>9765303>diverges because odd>>9765315>It has an over-harmonic majorant, so it convergesWhy do all of you idiots pretend to know more than you actually do?
>>9765324You're right, I'm an idiott. >>9765315
>>9765300>(standard for highschool students in countries where people poo in the loo)the funny thing is indians slaughter whites in math scores, that's why they're taking all your fucking jobs
>>9765300[eqn]\sum_{n\in\mathbb{Z}}f(n) = \frac{1}{2\pi i}\int_\mathbb{C}dz f(z)\csc(\pi z)[/eqn]
>>9765366We could easily fight back....if we valued cheating.
1/(n^3sin^2n)<=1/n^3 for all n. So yes, it converges.
>>9765377>1/(n^3sin^2n)<=1/n^3>[math]|\sin(x)|\leq 1 \implies \left|\frac{1}{\sin(x)}\right| \leq 1[/math]LMFAO
>>9765366they take jobs because they'll work for $2/hour you mongoloid
>>9765366Work with poos daily. Literal walking calculators, but literal calculators. Cannot under any circumstance extrapolate, conceptualize or apply calculations to ill-constrained equations (ie. reality). Must be spoon fed any and everything needed before "hard" math, results must be evaluated after
>>9765371>>9765383>>9765418bitter
>>9765307>just look up calc 2 to solve this unsolved problem
>>9765321Find out then idiot
>>9765369>infinity = infinity ok
>>9765300multiuply the whole thing by 1 to get rid of the numerator, and then it's simple from there
>>9765484>say some dumb shit>people respond to you>"hahahaha you must be all so stupid and bitter to reply to a retard like me"nice self worth
>>9765369f(n) is odd, so that trick won't work here.
>>9765300Wait isn't this an unsolved problem? I have seen this before. I believe it follows from the normality of pi.
>>9765300[math] \sin^2{(x)} [/math] will always be at most 1 so you can remove the term from the expression, henceforth its leaves you with [eqn] \displaystyle \sum_{n = 1}^\infty \frac{1}{n^3} = \Zeta (3) [/eqn]Q.E.D
>>9765300sum of 1/n^3 converges so if you multiply by a sine term it converges as well because sin(x) <= 1am I missing something?
>>9766025>>9766032The important terms are those where the sine is small since it's in the denominator.
[eqn]\forall n\,\in\,\mathbf N^*,\,0 \;<\; \frac1{n^3\, \left(\sin\,n\right)^2} \;=\; \underset{n\;\to\;\infty}O \left(\frac1{n^3}\right)[/eqn]and [math]\sum_{n\;\geqslant\;1}\frac1{n^3}[/math] is convergent, Q.E.D.Any more homework, OP?
>>9765300Probably converges, being an odd harmonic, even if I'm not even sure that [math]\frac{1}{sin^2(n)}[/math] is continuous. Should be for any integer value, since pi/2 and 3pi/2 aren't integers?>>9766054If you want some homework, try your hand at[math]\lim_{n\to \infty}\frac{sin(\frac{3}{n^n} + \frac{2}{n!})}{log(1+log(1-sin(\frac{5}{n!})))}[/math]
>>9766152Alright, actually give me a hand for this one tho[math]\displaystyle \lim_{n\to\infty}\frac{sin(\frac{3}{n^n} + \frac{2}{n!})}{log(1+log(1-sin(\frac{5}{n!})))}[/math]It is trivial to show that the denominator can be reduced to [math]\displaystyle \frac{5}{n!}[/math] using the known limits for [math]\displaystyle \lim_{n\to\infty}\frac{log(1+a_n)}{a_n}=1[/math] and [math]\displaystyle \lim_{n\to0} \frac{sin(a_n)}{a_n}=1[/math] with [math]a_n\to0[/math] but what the fuck do I do to the numerator? am I missing something?
>>9765300Lmfao there are actually people in this thread unironically using comparison test.
This thread has made me realize how retarded /sci is. I don't spend a lot of time on this board so I wasn't able to gauge the relative intelligence but dear fucking god. It approaches 0. You can literally just look at it quickly and see that. You don't need these retarded as long proofs. This is so basic that inspection and high school level math is literally all you need. I've literally lost all faith. Half of you need to unironically neck yourselves.
>>9766386you should leave then, this board is for intellectuals only
I love how /sci/'s true nature shows in threads like these. Post a seemingly simple problem and all the so called 150 IQ geniuses come out of the woodwork thinking "hey, I can do this", because they passed calc 1. Of course the retards are all wrong and even manage to overlook the posts that spell out for them it's an unsolved problem.
>>9765484cringe
>>9766386Motherfuck. 1/n approaches 0 too and does not converge as a series.
>>9766433that's the point of the threads like these
>>9766386Nice bait.
>>9765300Kek'd at all the brainlets in this thread who remove the sine because it's < 1.HEY IDIOTS THE WHOLE POINT IS THAT ITS VALUE CAN BE 0
>>9766494>ITS VALUE CAN BE 0No, it can't. The zeroes of the sine function are [math] \pi \mathbb{Z} [/math].Suppose you had an [math] n \in \mathbb{N} \setminus \{0\} [/math] and [math] k \in \mathbb{Z} [/math] with[eqn] n = \pi k [/eqn] then you could divide by [math] k [/math] to get[eqn] \pi = \frac{n}{k} \in \mathbb{Q} [/eqn]which is impossible since [math] \pi [/math] is irrational (the proof that [math] \pi [/math] is irrational has been left to the reader as an exercise).
>>9766494only if n is irrational, as sin is only 0 for [math]\frac{(2k+1)\pi}{2},k\in\mathbb N[/math]it can approach 0 for great numbers, but I don't think it'll ever be 0
>>9766518>>9766522But there could be values for which n^3(sinn)^2 < 1 for instance.
>>9766543such as?
>>9766550n = 3
>>9766550And furthermore n = 104347 nets us less than 7.
>>9766550>>9766572n=355 nets .04!
>>9766433There is a selection bias effect going on, though. The people that correctly note that they can't solve it, mostly don't post in the thread in the first place. Which makes the thread a very misleading place to gauge the relative proportions of those two groups.
>>9766594Yeah, well that's the internet at large. Still amusing though.
doesn't work with n=pi because sin(pi) is zero
>>9767406OP obviously will not recover from thid air tight argument.
>>9765300It really says a lot about the state of this sub that this post got so many unironic replies. I guess summer has begun.
>>9767459The biggest influx is during the winter though.
>>9765303>>9765315>>9765369>>9765377>>9766054
>>9765300I'm gonna go with yes, none of the values should be infinite, so the sum should be finite.
It's equal to 1 + 1/8 + .... + 1/n^3 at most, and that series converges.
>>9765300this is pretty obviousapply the steps of the extended euclidean to 1 and pias n approaches infinity, the general euclidean division step will produce a smaller remainder that is 61% ( [math]\frac{1}{\varphi}[/math] %) of the previous remainder's length. This approaches 0 exponentially, so it would be easy to find [math]n[/math] such that [math]n*1 - m*pi < \frac{1}{n\sqrt{n}}[/math]because sin(x) is close to x when x is close to a multiple of pi, we will always find n such that [eqn]\frac{1}{n^{3}sin(n)^{2}} > \frac{1}{n^{3}[\frac{1}{n\sqrt(n)}]^{2}}=1[/eqn] the general term doesn't approach 0, so the series does not converge
>>9765317nigger faggot
>>9767753*extended euclidean algorithm for real numbers
>>9765300Idk if it converges or not.Two things could be tried to prove divergence:1) show that for every e>0 there exists an n such that (n^3)sin(n)^2 < e.2) show that for some e>0 there exist infinitely many n such that (n^3)sin(n)^2 < e.Such a proof would probably involve understanding rational approximations of pi.
>>9765321>>9767753Flint Hill is a retard
>>9767753>as n approaches infinity, the general euclidean division step will produce a smaller remainder that is 61% ( 1/φ %) of the previous remainder's length. This approaches 0 exponentiallycitation?
>>9765321I'd lean towards no. A single instance of sin being closer to 0 than n^3 is larger than zero results in the series making a non-convergent jump in value. As n -> infinity, I don't see any evidence that these jumps in value would ever completely vanish.Though I guess that it would depend on exactly what pi is. If there's some tendency for multiples of pi to never get sufficiently close to whole numbers, then it would converge.
Was originally going to chastise OP for being a faggot and posting an open problem, but this thread has given me too many keks at this point. Not bad.
>>9765324>all of you idiots>idiot>some dumb shit>retarded>HEY IDIOTS>nigger faggotthis is the /sci/rrhoidal condition of /sci/
I play my trap card, having sin(n)^2 use degrees
The Answer is yes, because it gets simplified to Sonen∞
Shout out to Paa for the calculation help by the way
>>9767753Actually cringed. Like it's not just one of those photography-worthy cringes but this one is physical.
It's an unsolved problem, you're all retarded.
>>9766386has to be bait
>>9765300The sin term will never be 0, therefore no term will become infinife. All terms are therefore bounded. When n=inf, the term will be 0. Since all terms are bounded and converge to 0 contribution, the sum of the terms must be bounded and therefore the sum will converge. ¶
>>9765300https://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1
tell me a single theorem or invention you use on a day-to-day basis that came from India
>>9772385Im a big fan of butter chicken and naan.
>>9765300Nope it does not
>>9767599no it's not you retard
>>9765300it diverges in every p-adic field. proof: obvious
>>9772385STEM YouTube videos/lectures that are an absolute pain to watch due to the poo accent.
>>9772385curry
>>9772385Contributions of Bose, Mahalanobis, or Meghnad Saha come to mind.
>>9765377I needed a good laugh today.
>>9765300Converges
>>9767937>non-convergent jumpI wish to learn more about this concept.
>>9767599>It's equal to 1 + 1/8 + .... + 1/n^3 at most>at mostfunny way to spell "at least".
>>9772385zero/our digits/positional writing of numbers.algorithms for extracting square and cube rootsthe original rules of chess
>>9765300[math] \sum \dfrac{1}{n^2} > \sum \dfrac{1}{n^3(\sin{n})^2} [/math]Since [math]\sum \dfrac{1}{n^2} = \dfrac{\pi^2}{6}[/math], [math]\sum \dfrac{1}{n^3 (\sin{n})^2}[/math] must converge to a value lesser than that>mfw /sci/ hasn't passed calc II yet
>>9772385The [math] 1+2+3+...= - \dfrac{1}{12}[/math] shitpost
>>9776707>that first lineWhat?
>>9776707Did you even scroll down the thread a little bit before posting this? There was a link the the (previously open) problem with someone who gave a CORRECT explanation under it
>>9765300You don't own me, I don't have to solve this If don't want to.
Have all of you fucktarded autists seriously not just said p test p>1?Kys
>>9765300Seems convergent so yeah
Just use the ratio test. You can show that the series converges by the ratio test.
>>9776954It's not a p-series. If it was a p-series you would only have the n^3 term.
>>9765300[math]\sin(n)[/math] can't be zero because if [math] k \times \phi = \n [/math] for positive integer k and n then [math]\phi[/math] is rationalso [math]\frac {1}{\sin^2(n})>=M[/math] for some M and it's obvious that this series converges
>>9765300Direct comparison test. f(X) < Σ1/n^2, which converges, therefore it must also converge.
wait everyone heres actually retarded lol this is LITERALLY AN OPEN PROBLEM, ie LITERALLY NO MATHEMATICIAN IN THE WORLD HAS CRACKED IT YET. in fact its tied to the normality of pi, which has famously been known to be difficult as fuck to even begin proving. i fucking hate people who affirmatively say shit when they have literally no knowledge of the topic; it's insulting really