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anonymous
 5 years ago
summation of n=1 to infinty
ln((n^2+1)/(2n^2+2)) Does this diverge or converge?
anonymous
 5 years ago
summation of n=1 to infinty ln((n^2+1)/(2n^2+2)) Does this diverge or converge?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(2n ^{2}+2))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how does this converge?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(n^2+1)/(2(n^2+1)) = ln 1/2 is a finite number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, but for every n you add ln(1/2) And there are a lot of ns...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i had this one already. my question is rather this: \[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, I see you posted the problem again in equation form. Let me see

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes. the denominator in this case is n squared plus 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Newton, I have a doubt, \[\sum_{n=1}^{\infty}1/2 = ?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with the first one, the limit of the expression was 1/2 and the natural log was not equal to zero so it indeed converged. but i can't determine this one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that wouldn't make any sense since there is no n term in the summation. Isn't that right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It isn't summing a half once though, it is summing it for every n. n = 1 , add ln(1/2) n = 2 , add ln(1/2) n = .... etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you guys have a clue for the question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or should repost the question again? if you look above, you'd realize i posted a different but quite similar question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the first problem, it is infinite, for the second problem, as n reaches infinity, the summation reaches 0. So it is finite

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how? that's what i am not getting.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not the summation, the value of n^2+1/n^2/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh right, my mistake. Let me start afresh.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried the divergence test and it doesn't work. does it converge or diverge?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty}(n ^{2}+1)/(n ^{2}+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is the problem correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoops, its the natural log of that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for large values of n, \[n ^{2}+1/n ^{2}+2 \approx n ^{2}/n ^{2} =1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so as n approaches infinity, the natural log of the given equation approaches 0. so it is a finite number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0therefore it converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i did with the first question was to find the limit which was 1/2 and apply the natural log. With this question, what was your approach?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In general, the geometric series \[\sum_{n=0}^{\infty}z ^{n}\] converges if and only if z < 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you used the absolute convergence test i presume, which i am yet to learn. is that the only way out? can you outline the specifics of your step?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} \ln((n^2+1)/(2n^2+2))\] this sum does not converge

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think you're right since the ln 1/2 is not equal to zero for the divergence test, hence diverges.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you see the other question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are two ways to approach this problem. I am not sure which way is correct though. So here goes. \[(n ^{2}+1)/(n ^{2}+2) = (n ^{2}+21)/(n ^{2}+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dhatraditya you right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but we have natural logarithm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for large values of n, 1/n^2+2 approaches 0. so for large values of n, 1 1/n^2+2 = 10 = 1 ln 1 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it's not good i think we must do like as...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, ln 1 = 0 even for natural log. log 1 to any base is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty}\ln(n^2+1)\sum_{1}^{\infty}\ln(n^2+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you guys have me confused now. which approach is agreed upon now. and does it converge or not? can you also state the test you are using?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it converges. That is agreed upon. The method is what we are disagreeing upon.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i have learnt so far uses the various test methods like divergence test, integral test, basic comparison test, etc to solve this questions. which one are you applying in this case?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but this sums does not converge =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you talking about his sum or the previous one, hobbit?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because as far as I can tell, this sum seems to be converging.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but this sums does not converge =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i have some problems with charge on my mac)))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh well, then you should explain it to bbcram. because I seem to be wrong on this one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you don't understand bbcram?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so guys i have a question)) where you from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im from india Please explain how you got this sum is diverging.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool i am from Russia you must to know that ln(a/b)=ln(a)ln(b)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Russian eh? Cool. Student? yes ln(a/b)=lnalnb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and \[\sum_{?}^{?}(ab)=\sum_{?}^{?}a\sum_{?}^{?}b\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does Hobbit think: \[\sum_{1}^{\infty} \ln \frac{n^2+1}{n^2+2}\] DIVERGES? xD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, he seems to think so. I am asking him why he thinks that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0answer near 0.956448

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't worry Hobbit, Wolfram Alpha thinks so too ... xD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's not divergence, kiddo.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0INewton where you from? and where do you study?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry guys my pc suddenly restarted. It's no less clear now. INewton, dhatraditya, Hobbit. I am in the united states. I just started the topic on sequence and series this week and were challenged with this question. I have worked on others successfully but this is a headache. So what is your take? I seem to be seeing conflicting answers and approaches

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is a finite number. So it is convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then why did you think it was divergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont think that it is divergent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you said it does not converge.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0∑1∞ln(n2+1) is divergent and ∑1∞ln(n2+2) is divergent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so the ln of either the numerator or denominator in this case diverges. cool. however, the ln for the equation is 0 right? which makes it convergent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i don't know, but may be gamma function ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i in confuse too now =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think dhatraditya is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@dhatradtiya, can you just once again write the steps you took so i can write that down and study your procedure?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, \[(n ^{2}+1)/(n ^{2}+2)=((n ^{2}+2)−1))/(n ^{2}+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for large values of n,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1/(n ^{2}+2) \approx0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0therefore \[1−1/(n ^{2}+2) \approx10 = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[n \rightarrow \infty, \ln (11/(n ^{2}+2) = 0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for n=1 ln(11/3) for n=2 ln(11/6) for n=3 ln(11/11) for n=infty ln(10)=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but we must to find sum of first members

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(11/3)+ln(11/6)+ln(11/11)+...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It has been quite a while since I did my last convergence test. It seems obvious to me, but for proof, you will need to ask someone else. I can only give you my reasoning as to why it is convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks guys. are you mostly available on this forum?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I come here in my free time often to help out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool. can i give you my email or you give me yours. i could keep in touch

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm in university now =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you understand the reasoning, bbcram? It is not important to know the exact values of the series. You should be able to look at a series and tell if it is convergent or not directly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok cool what is you name? my email: galexant@yandex.ru

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my email id is dhatraditya@yahoo.com

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is your name bbcram?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm in the university too, except that with my calculus 11 professor essentially leaves everything to us. i would look over it carefully to understand the reasoning better and make good deductions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool. mine is rankbert@gmail.com. i use this more often.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dhatraditya you right thanks for discussion in discussion birth truth

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, it was a great discussion. Good luck to both of you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i like the way it all went. you guys are cool. thanks and stay in touch. i would stay in touch too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks good luck to you too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0goodluck to you both too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i would stay in touch

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@bbcram in what university do you study

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and what is you specialization?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0computer science. We can have further conversation through email if you don't mind. Thanks for your time and dhatraditya too. i'm out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool i am study on computer science too
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