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

- anonymous

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

- jamiebookeater

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- anonymous

it converges.

- anonymous

\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(2n ^{2}+2))\]

- anonymous

how does this converge?

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## More answers

- anonymous

ln(n^2+1)/(2(n^2+1)) = ln 1/2 is a finite number

- anonymous

Yes, but for every n you add ln(1/2)
And there are a lot of ns...

- anonymous

sorry i had this one already. my question is rather this:
\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]

- anonymous

wait, I see you posted the problem again in equation form. Let me see

- anonymous

yes. the denominator in this case is n squared plus 2

- anonymous

Newton, I have a doubt,
\[\sum_{n=1}^{\infty}1/2 = ?\]

- 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

that wouldn't make any sense since there is no n term in the summation. Isn't that right?

- 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

okay i understand.

- anonymous

do you guys have a clue for the question?

- anonymous

or should repost the question again? if you look above, you'd realize i posted a different but quite similar question.

- 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

how? that's what i am not getting.

- anonymous

not the summation, the value of n^2+1/n^2/2

- anonymous

it's n^2+1/n^2+2

- anonymous

oh right, my mistake. Let me start afresh.

- anonymous

i tried the divergence test and it doesn't work. does it converge or diverge?

- anonymous

\[\sum_{n=1}^{\infty}(n ^{2}+1)/(n ^{2}+2)\]

- anonymous

that is the problem correct?

- anonymous

whoops, its the natural log of that

- anonymous

yep

- anonymous

we know that ln 1 =0

- anonymous

for large values of n,
\[n ^{2}+1/n ^{2}+2 \approx n ^{2}/n ^{2} =1\]

- anonymous

so as n approaches infinity, the natural log of the given equation approaches 0. so it is a finite number

- anonymous

therefore it converges

- 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

In general, the geometric series
\[\sum_{n=0}^{\infty}z ^{n}\]
converges if and only if |z| < 1.

- 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

\[\sum_{1}^{\infty} \ln((n^2+1)/(2n^2+2))\] this sum does not converge

- anonymous

i think you're right since the ln 1/2 is not equal to zero for the divergence test, hence diverges.

- anonymous

do you see the other question?

- anonymous

what for example?

- anonymous

\[\sum_{1}^{\infty} \ln ((n ^{2}+1)/(n ^{2}+2))\]

- anonymous

this sum we can find

- anonymous

how?

- 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

dhatraditya
you right

- anonymous

\[=1-1/(n ^{2}+2)\]

- anonymous

yes

- anonymous

but we have natural logarithm

- 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

no it's not good
i think we must do like as...

- anonymous

yes, ln 1 = 0 even for natural log. log 1 to any base is 0.

- anonymous

\[\sum_{1}^{\infty}\ln(n^2+1)-\sum_{1}^{\infty}\ln(n^2+2)\]

- 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

it converges. That is agreed upon. The method is what we are disagreeing upon.

- 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

but this sums does not converge
=)

- anonymous

are you talking about his sum or the previous one, hobbit?

- anonymous

because as far as I can tell, this sum seems to be converging.

- anonymous

but this sums does not converge
=)

- anonymous

sorry i have some problems with charge on my mac)))

- anonymous

oh well, then you should explain it to bbcram. because I seem to be wrong on this one.

- anonymous

what do you don't understand bbcram?

- anonymous

so guys
i have a question))
where you from?

- anonymous

im from india
Please explain how you got this sum is diverging.

- anonymous

cool
i am from Russia
you must to know that ln(a/b)=ln(a)-ln(b)

- anonymous

Russian eh? Cool. Student?
yes ln(a/b)=lna-lnb

- anonymous

and \[\sum_{?}^{?}(a-b)=\sum_{?}^{?}a-\sum_{?}^{?}b\]

- anonymous

Does Hobbit think:
\[\sum_{1}^{\infty} \ln \frac{n^2+1}{n^2+2}\]
DIVERGES? xD

- anonymous

yes

- anonymous

yes, he seems to think so. I am asking him why he thinks that

- anonymous

answer near -0.956448

- anonymous

huh?

- anonymous

don't worry Hobbit, Wolfram |Alpha thinks so too ... xD

- anonymous

That's not divergence, kiddo.

- anonymous

it s partial sum

- anonymous

INewton where you from?
and where do you study?

- 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

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

that is a finite number. So it is convergent.

- anonymous

yes

- anonymous

Then why did you think it was divergent?

- anonymous

i dont think that it is divergent

- anonymous

you said it does not converge.

- anonymous

âˆ‘1âˆžln(n2+1) is divergent
and âˆ‘1âˆžln(n2+2) is divergent

- anonymous

is not it?

- anonymous

yes.

- 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

i don't know, but may be gamma function ?

- anonymous

i in confuse too now =)

- anonymous

i think dhatraditya is right.

- anonymous

@dhatradtiya, can you just once again write the steps you took so i can write that down and study your procedure?

- anonymous

yes,
\[(n ^{2}+1)/(n ^{2}+2)=((n ^{2}+2)âˆ’1))/(n ^{2}+2)\]

- anonymous

\[=1âˆ’1/(n ^{2}+2)\]

- anonymous

for large values of n,

- anonymous

\[1/(n ^{2}+2) \approx0\]

- anonymous

yes and so

- anonymous

therefore \[1âˆ’1/(n ^{2}+2) \approx1-0 = 1\]

- anonymous

but 1-1=0 not 1

- anonymous

\[n \rightarrow \infty, \ln (1-1/(n ^{2}+2) = 0\]

- 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

yes
but we must to find sum of first members

- anonymous

ln(1-1/3)+ln(1-1/6)+ln(1-1/11)+...

- 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

thanks guys. are you mostly available on this forum?

- anonymous

I come here in my free time often to help out.

- anonymous

cool. can i give you my email or you give me yours. i could keep in touch

- anonymous

i'm in university now =)

- 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

ok
cool
what is you name?
my e-mail: galexant@yandex.ru

- anonymous

my email id is dhatraditya@yahoo.com

- anonymous

what is your name bbcram?

- 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

cool. mine is rankbert@gmail.com. i use this more often.

- anonymous

dhatraditya you right
thanks for discussion
in discussion birth truth

- anonymous

yes, it was a great discussion. Good luck to both of you!

- anonymous

i like the way it all went. you guys are cool. thanks and stay in touch. i would stay in touch too.

- anonymous

thanks good luck to you too

- anonymous

goodluck to you both too

- anonymous

yes i would stay in touch

- anonymous

@bbcram in what university do you study

- anonymous

CUNY in New York

- anonymous

and what is you specialization?

- 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

ok ciao guys.

- anonymous

cool
i am study on computer science too

- anonymous

ciao

- anonymous

see soon here

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