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The hard sums/products thread.

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Original post by Gregorius
I think it's just to keep control over where the negative numbers appear and to give a standard form to the CF. Otherwise you would get multiple ways of expressing the same CF.


I think I've had an idea, I've proven that xn+1=1−1xnx_{n+1} = 1 - \frac{1}{x_n} is periodic with period 3, does that mean it only converges if it's constant? My gut feeling is to say yes but I know I've been mislead by periodic sequences and constant functions in the past so probably not...
Original post by Zacken
Staring at that for a while got me thinking that really, when we're evaluating continued fractions, what we're really doing is assuming some limit ℓ\ell exists and then using the recursive sequence since ℓ\ell satisfies ℓ=1−1ℓ\ell = 1 - \frac{1}{\ell} - is that line of reasoning correct?


Yes. Much like those silly proofs along the lines of "assuming S = Σ 1 exists then S = -1/2"

(before someone says something about analytic continuation, yes one can make sense of this result, but not in the manner that is often depicted on the internet)

So I reckon that I've proved that xn+1=1−1xnx_{n+1} = 1 - \frac{1}{x_n} is a periodic sequence with period 33 (have I actually? Or do I need something more general than just checking those first few terms?), but I've not really got a clue where to go from here.


Indeed you have, just replace 0 with n. Something not trivial though: is it a coincidence that there is also a 33 in the solutions to the equation you solved, and, if not, what is the connection? :wink:
Original post by Lord of the Flies
Yes. Much like those silly proofs along the lines of "assuming S = Σ 1 exists then S = -1/2"


Ah, okay. Thanks.

Indeed you have, just replace 0 with n. Something not trivial though: is it a coincidence that there is also a 33 in the solutions to the equation you solved, and, if not, what is the connection? :wink:


...oh dear, I hadn't even thought there would be a link. Would it happen to have anything to do with trigonometrical functions? I shall report back after some thought. :biggrin:
Original post by Zacken

Is there a particular reason for this convention? I don't think it quite helps here, as far as I can see. :colondollar:


So if you put your continued fraction in the "standard form" and then define the two sequences

h−2=0,h−1=1,k−2=1,k−1=0 \displaystyle h_{-2} = 0, h_{-1}=1, k_{-2}=1, k_{-1}=0
hn=anhn−1+hn−2 \displaystyle h_{n} = a_{n} h_{n-1} + h_{n-2}
kn=ankn−1+kn−2 \displaystyle k_{n} = a_{n} k_{n-1} + k_{n-2}

then it is standard theory that if the continued fraction converges, it converges to the limit of hn/knh_n/k_n as n tends to infinity. You might like to try this on your CF!
Admittedly a little on the easy side, but cute nonetheless:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\zeta(2)}{2} + \frac{\zeta(4)}{2^3} + \frac{\zeta(6)}{2^5} + \frac{\zeta(8)}{2^7} + \cdots \end{equation*}

Reply 165
Avatar for RichE
RichE
15
Show that

[br][br]∑n=0∞1F2n+1+1=52[br][br][br][br]\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}+1} = \frac{\sqrt{5}}{2}[br][br]

where Fn F_{n} is the nth Fibonacci number.
Reply 166
Avatar for RichE
RichE
15
Show that

[br][br]∑n=0∞1F2n=7−52[br][br][br][br]\sum_{n=0}^{\infty} \frac{1}{F_{2^n}} = \frac{7 - \sqrt{5}}{2}[br][br]

where Fn F_{n} is the nth Fibonacci number.
Original post by Zacken
Admittedly a little on the easy side, but cute nonetheless:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\zeta(2)}{2} + \frac{\zeta(4)}{2^3} + \frac{\zeta(6)}{2^5} + \frac{\zeta(8)}{2^7} + \cdots \end{equation*}



∑k=1∞ζ(2k)22k−1[br][br]=2∑k=1∞∑n=1∞122kn2k[br][br]=2∑n=1∞14n2−1(∗)[br][br]=1 \displaystyle\sum_{k=1}^{ \infty} \dfrac{ \zeta (2k)}{2^{2k-1}} [br][br]= 2 \displaystyle\sum_{k=1}^{ \infty} \displaystyle\sum_{n=1}^{ \infty} \dfrac{1}{2^{2k} n^{2k}} [br][br]= 2 \displaystyle\sum_{n=1}^{ \infty} \dfrac{1}{4n^2 -1} (*)[br][br]= 1

Without some justification for the limit swapping to get to (∗) (*) I am no more than an engineer: expand by definitions and it swapping the limits should work for anything which is finite when evaluated, by taking the limits in different orders.
Original post by RichE
Show that

[br][br]∑n=0∞1F2n=7−52[br][br][br][br]\sum_{n=0}^{\infty} \frac{1}{F_{2^n}} = \frac{7 - \sqrt{5}}{2}[br][br]

where Fn F_{n} is the nth Fibonacci number.


Original post by RichE
Show that

[br][br]∑n=0∞1F2n+1+1=52[br][br][br][br]\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}+1} = \frac{\sqrt{5}}{2}[br][br]

where Fn F_{n} is the nth Fibonacci number.


the usual way is to either rearrange to get the summation interms of itself or just sub in explicit expression for Fn F_n and get geometric

REEË
Original post by EnglishMuon
the usual way is to either rearrange to get the summation interms of itself or just sub in explicit expression for Fn F_n and get geometricBrute force with the explict formula for F_n works for the F2n+1F_{2n+1} one but doesn't seem to come out nicely for the other.

For the F2nF_{2^n} one, I suggest working out a few terms expliclty, and then form a hypothesis for what ∑k=0n1F2k\displaystyle \sum_{k=0}^n \dfrac{1}{F_{2^k}} is. You can then prove this correct by induction (and then take the limit).
Original post by RichE
Show that

∑n=0∞1F2n=7−52[br][br]\displaystyle \sum_{n=0}^{\infty} \frac{1}{F_{2^n}} = \frac{7 - \sqrt{5}}{2}[br][br]

where Fn F_{n} is the nth Fibonacci number.


Using Binet we get

Unparseable latex formula:

\displaystyle [br]\begin{align*} \sum_{n> 0} \frac{\sqrt{5}}{\phi^{2^n} - \phi^{-2^n}} &= \sqrt{5}\sum_{n> 0} \frac{\phi^{2^n}}{\left(\phi^{2^n}\right)^2 - 1} \\ & = \sqrt{5} \sum_{n> 0} \frac{1}{\phi^{2^n} - 1} - \frac{1}{\phi^{2^{n+1}} - 1} \\ & = \frac{\sqrt{5}}{\phi^2- 1}\end{align*}



where the penultimate equality follows from telecoping.

Hence the required sum is 1+5ϕ=7−521 + \frac{\sqrt{5}}{\phi} = \frac{7-\sqrt{5}}{2}

The induction approach suggested above is also nice and works well here, but isn't necessary for the Fibonnaci sequence. It does however, come in useful when one wants to prove a similar result for ∑n≥0a2n−1\sum_{n \geq 0}a_{2^n}^{-1} where ana_n follows a generalised Fibonnaci recurrence.
(edited 6 years ago)
Original post by Zacken
Using Binet we get

Unparseable latex formula:

\displaystyle [br]\begin{align*} \sum_{n\geq 0} \frac{\sqrt{5}}{\phi^{2^n} - \phi^{-2^n}} &= \sqrt{5}\sum_{n\geq 0} \frac{\phi^{2^n}}{\left(\phi^{2^n}\right)^2 - 1} \\ & = \sqrt{5} \sum_{n\geq 0} \frac{1}{\phi^{2^n} - 1} - \frac{1}{\phi^{2^{n+1}} - 1} \\ & = \frac{\sqrt{5}}{\phi - 1}= \frac{1}{2} \left(7-\sqrt{5} \right) \end{align*}



where the penultimate equality follows from telecoping.

The induction approach suggested above is also nice and works well here, but isn't necessary for the Fibonnaci sequence. It does however, come in useful when one wants to prove a similar result for ∑n≥0a2n−1\sum_{n \geq 0}a_{2^n}^{-1} where ana_n follows a generalised Fibonnaci recurrence.


Did you really USE BINET????? Did you sit him down by ur butch annies, give him an outofdate cookie and say "Eyy Binet, help me evaluate this sum", "sure thang @Zacken ". @Clement Mouhot,
Original post by Zacken
Using Binet we get

Unparseable latex formula:

\displaystyle [br]\begin{align*} \sum_{n> 0} \frac{\sqrt{5}}{\phi^{2^n} - \phi^{-2^n}} &= \sqrt{5}\sum_{n> 0} \frac{\phi^{2^n}}{\left(\phi^{2^n}\right)^2 - 1} \\ & = \sqrt{5} \sum_{n> 0} \frac{1}{\phi^{2^n} - 1} - \frac{1}{\phi^{2^{n+1}} - 1} \\ & = \frac{\sqrt{5}}{\phi^2- 1}\end{align*}



where the penultimate equality follows from telecoping.I confess, I'm not seeing where this penultimate equality comes from...
Original post by DFranklin
I confess, I'm not seeing where this penultimate equality comes from...


Which is the penultimate equality - the partial fractions bit? If so, I'm struggling with that myself.

I got as far as that earlier today, but couldn't see how it would split up nicely.
Original post by atsruser
Which is the penultimate equality - the partial fractions bit? If so, I'm struggling with that myself.

I got as far as that earlier today, but couldn't see how it would split up nicely.
Yeah, I partial fractioned, but got ϕ2n±1\phi^{2^n} \pm 1 for the denominators (up to a constant).and I couldn't see any obvious way to progress...
Can someone help me with this:
https://www.thestudentroom.co.uk/showthread.php?p=71370910&highlight=
Original post by Laughh3
Please don't make OT posts to other threads in an effort to get people to help you. It generally doesn't work very well...
Original post by DFranklin
Please don't make OT posts to other threads in an effort to get people to help you. It generally doesn't work very well...


Okay...
Original post by DFranklin
I confess, I'm not seeing where this penultimate equality comes from...


Original post by atsruser
Which is the penultimate equality - the partial fractions bit? If so, I'm struggling with that myself.

I got as far as that earlier today, but couldn't see how it would split up nicely.


Surely if you let x=ϕ2nx=\phi^{2^n} you have 1x−x−1=xx2−1=1x−1−1x2−1\displaystyle \frac{1}{x - x^{-1}} = \frac{x}{x^2 - 1} = \frac{1}{x-1} - \frac{1}{x^2 - 1}.

Edit: okay, perhaps not partial fractions per se.
(edited 6 years ago)
Original post by Zacken
Surely if you let x=ϕ2nx=\phi^{2^n} you have 1x−x−1=xx2−1=1x−1−1x2−1\displaystyle \frac{1}{x - x^{-1}} = \frac{x}{x^2 - 1} = \frac{1}{x-1} - \frac{1}{x^2 - 1}.
Nice. I went the 1/(x-1) + 1/(x+1) route and got stuck. The mass of exponents confused me too...

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