anonymous
  • anonymous
summation of n=1 to infinty ln((n^2+1)/(2n^2+2)) Does this diverge or converge?
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
it converges.
anonymous
  • anonymous
\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(2n ^{2}+2))\]
anonymous
  • anonymous
how does this converge?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ln(n^2+1)/(2(n^2+1)) = ln 1/2 is a finite number
anonymous
  • anonymous
Yes, but for every n you add ln(1/2) And there are a lot of ns...
anonymous
  • anonymous
sorry i had this one already. my question is rather this: \[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]
anonymous
  • anonymous
wait, I see you posted the problem again in equation form. Let me see
anonymous
  • anonymous
yes. the denominator in this case is n squared plus 2
anonymous
  • anonymous
Newton, I have a doubt, \[\sum_{n=1}^{\infty}1/2 = ?\]
anonymous
  • anonymous
with 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
  • anonymous
that wouldn't make any sense since there is no n term in the summation. Isn't that right?
anonymous
  • anonymous
It 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
  • anonymous
okay i understand.
anonymous
  • anonymous
do you guys have a clue for the question?
anonymous
  • anonymous
or should repost the question again? if you look above, you'd realize i posted a different but quite similar question.
anonymous
  • anonymous
for the first problem, it is infinite, for the second problem, as n reaches infinity, the summation reaches 0. So it is finite
anonymous
  • anonymous
how? that's what i am not getting.
anonymous
  • anonymous
not the summation, the value of n^2+1/n^2/2
anonymous
  • anonymous
it's n^2+1/n^2+2
anonymous
  • anonymous
oh right, my mistake. Let me start afresh.
anonymous
  • anonymous
i tried the divergence test and it doesn't work. does it converge or diverge?
anonymous
  • anonymous
\[\sum_{n=1}^{\infty}(n ^{2}+1)/(n ^{2}+2)\]
anonymous
  • anonymous
that is the problem correct?
anonymous
  • anonymous
whoops, its the natural log of that
anonymous
  • anonymous
yep
anonymous
  • anonymous
we know that ln 1 =0
anonymous
  • anonymous
for large values of n, \[n ^{2}+1/n ^{2}+2 \approx n ^{2}/n ^{2} =1\]
anonymous
  • anonymous
so as n approaches infinity, the natural log of the given equation approaches 0. so it is a finite number
anonymous
  • anonymous
therefore it converges
anonymous
  • anonymous
what 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
  • anonymous
In general, the geometric series \[\sum_{n=0}^{\infty}z ^{n}\] converges if and only if |z| < 1.
anonymous
  • anonymous
so 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
  • anonymous
\[\sum_{1}^{\infty} \ln((n^2+1)/(2n^2+2))\] this sum does not converge
anonymous
  • anonymous
i think you're right since the ln 1/2 is not equal to zero for the divergence test, hence diverges.
anonymous
  • anonymous
do you see the other question?
anonymous
  • anonymous
what for example?
anonymous
  • anonymous
\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]
anonymous
  • anonymous
this sum we can find
anonymous
  • anonymous
how?
anonymous
  • anonymous
there are two ways to approach this problem. I am not sure which way is correct though. So here goes. \[(n ^{2}+1)/(n ^{2}+2) = (n ^{2}+2-1)/(n ^{2}+2)\]
anonymous
  • anonymous
dhatraditya you right
anonymous
  • anonymous
\[=1-1/(n ^{2}+2)\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
but we have natural logarithm
anonymous
  • anonymous
for large values of n, 1/n^2+2 approaches 0. so for large values of n, 1- 1/n^2+2 = 1-0 = 1 ln 1 = 0
anonymous
  • anonymous
no it's not good i think we must do like as...
anonymous
  • anonymous
yes, ln 1 = 0 even for natural log. log 1 to any base is 0.
anonymous
  • anonymous
\[\sum_{1}^{\infty}\ln(n^2+1)-\sum_{1}^{\infty}\ln(n^2+2)\]
anonymous
  • anonymous
you 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
  • anonymous
it converges. That is agreed upon. The method is what we are disagreeing upon.
anonymous
  • anonymous
what 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
  • anonymous
but this sums does not converge =)
anonymous
  • anonymous
are you talking about his sum or the previous one, hobbit?
anonymous
  • anonymous
because as far as I can tell, this sum seems to be converging.
anonymous
  • anonymous
but this sums does not converge =)
anonymous
  • anonymous
sorry i have some problems with charge on my mac)))
anonymous
  • anonymous
oh well, then you should explain it to bbcram. because I seem to be wrong on this one.
anonymous
  • anonymous
what do you don't understand bbcram?
anonymous
  • anonymous
so guys i have a question)) where you from?
anonymous
  • anonymous
im from india Please explain how you got this sum is diverging.
anonymous
  • anonymous
cool i am from Russia you must to know that ln(a/b)=ln(a)-ln(b)
anonymous
  • anonymous
Russian eh? Cool. Student? yes ln(a/b)=lna-lnb
anonymous
  • anonymous
and \[\sum_{?}^{?}(a-b)=\sum_{?}^{?}a-\sum_{?}^{?}b\]
anonymous
  • anonymous
Does Hobbit think: \[\sum_{1}^{\infty} \ln \frac{n^2+1}{n^2+2}\] DIVERGES? xD
anonymous
  • anonymous
yes
anonymous
  • anonymous
yes, he seems to think so. I am asking him why he thinks that
anonymous
  • anonymous
answer near -0.956448
anonymous
  • anonymous
huh?
anonymous
  • anonymous
don't worry Hobbit, Wolfram |Alpha thinks so too ... xD
anonymous
  • anonymous
That's not divergence, kiddo.
anonymous
  • anonymous
it s partial sum
anonymous
  • anonymous
INewton where you from? and where do you study?
anonymous
  • anonymous
sorry 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
  • anonymous
http://www.wolframalpha.com/input/?i=sum+ln+%28%28%28j^2%29%2B1%29%2F%28j^2%2B2%29%29%2C+j%3D1+to+infinity
anonymous
  • anonymous
that is a finite number. So it is convergent.
anonymous
  • anonymous
yes
anonymous
  • anonymous
Then why did you think it was divergent?
anonymous
  • anonymous
i dont think that it is divergent
anonymous
  • anonymous
you said it does not converge.
anonymous
  • anonymous
∑1∞ln(n2+1) is divergent and ∑1∞ln(n2+2) is divergent
anonymous
  • anonymous
is not it?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
ok, 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
  • anonymous
i don't know, but may be gamma function ?
anonymous
  • anonymous
i in confuse too now =)
anonymous
  • anonymous
i think dhatraditya is right.
anonymous
  • anonymous
@dhatradtiya, can you just once again write the steps you took so i can write that down and study your procedure?
anonymous
  • anonymous
yes, \[(n ^{2}+1)/(n ^{2}+2)=((n ^{2}+2)−1))/(n ^{2}+2)\]
anonymous
  • anonymous
\[=1−1/(n ^{2}+2)\]
anonymous
  • anonymous
for large values of n,
anonymous
  • anonymous
\[1/(n ^{2}+2) \approx0\]
anonymous
  • anonymous
yes and so
anonymous
  • anonymous
therefore \[1−1/(n ^{2}+2) \approx1-0 = 1\]
anonymous
  • anonymous
but 1-1=0 not 1
anonymous
  • anonymous
\[n \rightarrow \infty, \ln (1-1/(n ^{2}+2) = 0\]
anonymous
  • anonymous
for n=1 ln(1-1/3) for n=2 ln(1-1/6) for n=3 ln(1-1/11) for n=infty ln(1-0)=0
anonymous
  • anonymous
yes but we must to find sum of first members
anonymous
  • anonymous
ln(1-1/3)+ln(1-1/6)+ln(1-1/11)+...
anonymous
  • anonymous
It 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
  • anonymous
thanks guys. are you mostly available on this forum?
anonymous
  • anonymous
I come here in my free time often to help out.
anonymous
  • anonymous
cool. can i give you my email or you give me yours. i could keep in touch
anonymous
  • anonymous
i'm in university now =)
anonymous
  • anonymous
did 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
  • anonymous
ok cool what is you name? my e-mail: galexant@yandex.ru
anonymous
  • anonymous
my email id is dhatraditya@yahoo.com
anonymous
  • anonymous
what is your name bbcram?
anonymous
  • anonymous
i'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
  • anonymous
cool. mine is rankbert@gmail.com. i use this more often.
anonymous
  • anonymous
dhatraditya you right thanks for discussion in discussion birth truth
anonymous
  • anonymous
yes, it was a great discussion. Good luck to both of you!
anonymous
  • anonymous
i like the way it all went. you guys are cool. thanks and stay in touch. i would stay in touch too.
anonymous
  • anonymous
thanks good luck to you too
anonymous
  • anonymous
goodluck to you both too
anonymous
  • anonymous
yes i would stay in touch
anonymous
  • anonymous
@bbcram in what university do you study
anonymous
  • anonymous
CUNY in New York
anonymous
  • anonymous
and what is you specialization?
anonymous
  • anonymous
computer science. We can have further conversation through email if you don't mind. Thanks for your time and dhatraditya too. i'm out
anonymous
  • anonymous
ok ciao guys.
anonymous
  • anonymous
cool i am study on computer science too
anonymous
  • anonymous
ciao
anonymous
  • anonymous
see soon here

Looking for something else?

Not the answer you are looking for? Search for more explanations.