Question

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

Answer 1

it converges.

Answer 2

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

Answer 3

how does this converge?

Answer 4

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

Answer 5

Yes, but for every n you add ln(1/2)

And there are a lot of ns...

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

okay i understand.

Answer 14

do you guys have a clue for the question?

Answer 15

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

Answer 16

for the first problem, it is infinite, for the second problem, as n reaches infinity, the summation reaches 0. So it is finite

Answer 17

how? that's what i am not getting.

Answer 18

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

Answer 19

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

Answer 20

oh right, my mistake. Let me start afresh.

Answer 21

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

Answer 22

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

Answer 23

that is the problem correct?

Answer 24

whoops, its the natural log of that

Answer 25

yep

Answer 26

we know that ln 1 =0

Answer 27

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

Answer 28

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

Answer 29

therefore it converges

Answer 30

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?

Answer 31

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

Answer 32

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?

Answer 33

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

Answer 34

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

Answer 35

do you see the other question?

Answer 36

what for example?

Answer 37

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

Answer 38

this sum we can find

Answer 39

how?

Answer 40

there are two ways to approach this problem. I am not sure which way is correct though. So here goes.

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

Answer 41

dhatraditya
you right

Answer 42

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

Answer 43

yes

Answer 44

but we have natural logarithm

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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?

Answer 50

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

Answer 51

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?

Answer 52

but this sums does not converge
=)

Answer 53

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

Answer 54

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

Answer 55

but this sums does not converge
=)

Answer 56

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

Answer 57

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

Answer 58

what do you don't understand bbcram?

Answer 59

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

Answer 60

im from india

Please explain how you got this sum is diverging.

Answer 61

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

Answer 62

Russian eh? Cool. Student?

yes ln(a/b)=lna-lnb

Answer 63

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

Answer 64

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

Answer 65

yes

Answer 66

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

Answer 67

answer near -0.956448

Answer 68

huh?

Answer 69

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

Answer 70

That's not divergence, kiddo.

Answer 71

it s partial sum

Answer 72

INewton where you from?
and where do you study?

Answer 73

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

Answer 74

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

Answer 75

that is a finite number. So it is convergent.

Answer 76

yes

Answer 77

Then why did you think it was divergent?

Answer 78

i dont think that it is divergent

Answer 79

you said it does not converge.

Answer 80

∑1∞ln(n2+1) is divergent
and ∑1∞ln(n2+2) is divergent

Answer 81

is not it?

Answer 82

yes.

Answer 83

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?

Answer 84

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

Answer 85

i in confuse too now =)

Answer 86

i think dhatraditya is right.

Answer 87

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

Answer 88

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

Answer 89

\[=1−1/(n ^{2}+2)\]

Answer 90

for large values of n,

Answer 91

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

Answer 92

yes and so

Answer 93

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

Answer 94

but 1-1=0 not 1

Answer 95

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

Answer 96

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

Answer 97

yes
but we must to find sum of first members

Answer 98

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