General

[quote=Invictinite]f(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n *sqrt(5))
where n is the fibonacci number that you want

so looking at your series
(5/(1*2)) + (5/(2*3)) + (5/(3*5))

first term is the third term in fibonacci so you would use n+2 for all the n's in my equation.

so basically:
5 / (n * f(n + 2)) and f(n) is defined above.[/quote]
GENIUS OMGG Thank you!

f(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n *sqrt(5))
where n is the fibonacci number that you want

so looking at your series
(5/(1*2)) + (5/(2*3)) + (5/(3*5))

first term is the third term in fibonacci so you would use n+2 for all the n's in my equation.

so basically:
5 / (n * f(n + 2)) and f(n) is defined above.

@BadCereal: I think there's a special notation for fibonacci numbers... can't remember it exactly and pretty sure it involved using combination.

For the sake of your question, I'm not certain to what detail you need to come up with an answer.. because by simply looking at it you know that it's 5*(1/2+1/6+1/15+....). Given that 1/6 is less than half of 1/2, you know that if we continue this sequence it'll be less than 1. I.e. The answer should be approaching 5.

[quote=BadCereal]But how do you do it when the series goes to infinite? I don't think combinations apply here.
@QuantumLegend: Never heard of golden ratios so I just wikied it now. Makes sense but how do you suppose in making a general statement?
That's the hard part[/quote]

Yea the hard part is getting it into a general equation...the thing is that each number of the golden ratio relates to the two previous numbers. You would have like an individual expression for the denominator..or...something...
:/

[quote=ryuushinou]@badcereal, @quantumlegend

Oh woops.. totally forgot about the first..

Probably need to use nCr then...[/quote]
But how do you do it when the series goes to infinite? I don't think combinations apply here.
@QuantumLegend: Never heard of golden ratios so I just wikied it now. Makes sense but how do you suppose in making a general statement?
That's the hard part

The closest rational approximations to the golden ratio are 2/1, 3/2, 5/3, 8/5, the inverses of which fit your denominator.

edit: Just realized I make no sense when I say the inverse of them.

@badcereal, @quantumlegend

Oh woops.. totally forgot about the first..
Probably need to use nCr then... or Fn...notation

What grade maths is this? I can't even remember what I did for uni...

Anyway, the answer should be approaching 5.

[quote=Copyrighted]I think you're missing half the problem. Typically they list the first 3 terms then the general form. As for which convergence test to use, it's commonly ratio test so go with that first if you are completely clueless. Integral test if it's easy to integrate I guess.[/quote]

The problem literally states only this (word for word):
Find the sum of the following infinite series: And then the 3 terms are all they give.
Man this sucks

[quote=QuantumLegend]Well the pattern seems to be (5/(n*(n+1))).
Idk what a convergence test is though, I'm in BC, so I would just be using the summation formula.
/maybe that doesn't work with infinites, i never remember[/quote]

that's not the equation since in the 3 term the n+1 would have to be equal to 5 but it's 4.
that "n + 1" part looks more like fibbonacci. you probably need to use the golden ratio in order to model it.

[quote=ryuushinou]Is that
5 _
(n*(2n-1))[/quote]

No, because that doesn't work for the first term.
Mine was wrong too.
That second term in the denominator follows the Fibonacci pattern starting from the third term actually. Though idk how to express that.

[quote=ryuushinou]Is that
5 _
(n*(2n-1))[/quote]
If it was then the first term would be (5/(1*1)).

Is that
5 _
(n*(2n-1))

[quote=QuantumLegend]Well the pattern seems to be (5/(n*(n+1))).
Idk what a convergence test is though, I'm in BC, so I would just be using the summation formula.
/maybe that doesn't work with infinites, i never remember[/quote]
It's not (5/(n*(n+1)), look at the third term.
If it was that, the third term would be (5/(3*4))

Well the pattern seems to be (5/(n*(n+1))).
Idk what a convergence test is though, I'm in BC, so I would just be using the summation formula.
/maybe that doesn't work with infinites, i never remember

[quote=Erag0n1]Isn't that pre-calculus?[/quote]
[url=http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx]No you need to apply convergence tests to this problem because the series is infinite.[/url]
Edit: Whoops, wrong section. Mod change it pls.

Isn't that pre-calculus?

wolframalpha

/endthread