Printable View
What is the sum to infinity of this infinite series?:
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + ...
There's Fibonacci numbers for the numerator and powers of two for the denonimator.
I want a method and an answer so you can't stick it in Excel!
The limit seems to be 3.
Do you have any reason to believe that there is a method of proving what this limit is?
Hi
I would think the limit would be 2. The first 3 add to 1 and the next zillion are never going to exceed one becos the denom rises in multiples whereas the numerator rises in adds. Anyway, i gotta go zzzzzzzzzz cos is too late to think :D
Regards
Stuart
the limit is actually 2.
it could be proved with binnet's formula or more easily...
i'll post the easier way tomorrow.
Sorry to spoil your idea.
I can prove that the sum of the series is greater than 3
The given series
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + ... ----->(1)
Now let us consider this part of the series...
1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + ... ------->(2)
Also consider the following Aritmetico-Geometrico series,
1/4 + 2/8 + 3/16 + 4/32 + 5/64 + 6/128 + ... ------->(3)
With common difference d = 1, common ratio r = 1/2 and
the first term of the sequence a = 1/4
The sum to Infinty of this Aritmetico-Geometrico Series, is
given by
sum to infinity = [ a/(1-r) ] + [ d*r/(1-r)^2 ]
{If you need the derivation for this formula, I'll post it later or
see http://www.efunda.com/math/seriesofc..._GeoSeries.cfm }
so according to this,
sum to infinity of aritmetico-geometrico series
SAG = [ (1/4) / (1-1/2) ] + [ (1*1/2)/(1-1/2)^2 ]
SAG = 1/2 + 2 = 2.5
now compare the corresponding terms of the series (2) and the
AG series (3),
1/4 = 1/4 , 2/8 = 2/8 , 3/16 = 3/16 and then...
5/32 > 4/32 , 8/64 > 5/64 , 13/128 > 6/128 and so on...
In short..
1/4 + 2/8 + 3/16 + 5/32 + 8/64 + ... > 1/4 + 2/8 + 3/16 + 4/32 + 5/64 + ....
and we know the sum on the right side = 2.5
so, 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + ... > 2.5
so the original series (1)
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + ... > 3
:)
I use Jamie's formula to dispute your's
Your answer = b ^ 0 + 11 - 0 * X :D It's 2 !!!
BeachBum: You are blasphemous and guilty of lese majesty! You have accused me of making a mistake. Do you not know that I am infallible? It should be considered illegal, unethical, fattening, and without any socially redeeming features to accuse me of being in error.
The fact that I made an error does not excuse your heinous behavior. So I said that the limit was Three, not two. That is not much of an error, only 50% too big. If I spend some time on this, I can probably invent an excuse for saying 3 instead of 2, and find some weird way of proving that you, not I made a mistake. I do not have time for that project right now. Hold your breath until I can work on it.
DavidHooper: You are a coconspirator and just as guilty as BeachBum.
ThinkTank2: Check your application of Arithemetico-Geometrico series. If first term is 1/4, second term is (1 + 1/4)/8, not 2/8.
BTW: Thanx for mentioning the site. It looks interesting. Either I never heard of or forgot about Arithemetico-Geometrico series. Interesting concept & formula.
As for the problem of proving the limit of this series. Some experimentation strongly indicates that the limit is 2, as previously posted by Hooper and the Bum.
A possible approach to proving that the limit is two, is to find two series which bound this series and which both have two as a limit. I fooled with several series which have two as a limit and which are always greater that the sum of the series being analyzed. I have not yet found a sereis for use as a lower bound.
http://www.vbforums.com/images/ieimages/2001/11/14.gifhttp://www.vbforums.com/images/ieimages/2001/11/15.gif
consider this AG series,
-1 + 0/2 + 1/4 + 2/8 + 3/16 + 4/32 + 5/64 + 6/128 +...
a = -1, d=1 , r=1/2
The sum to Infinity of the AG series : [-1/(1-1/2)] + [(1*1/2)/(1-1/2)^2]
= -2 + 2 = 0
That Implies,
1/4 + 2/8 + 3/16 + 4/32 + 5/64 + 6/128 +... = 1
Comparing this with (2)
1/4 + 2/8 + 3/16 + 5/32 + 8/64 + ... > 1
So the original series
1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + ... > 1.5
hmmm... not very helpful.
bounding the series is neat - i've never seen that done before. unfortunately it's quite tricky to find the right series.
a boring way is to use binnet's formula (http://www.mcs.surrey.ac.uk/Personal...a.html#formula)
for the nth term. express the formula in partial fractions and then sum to infinity by the method of differencing (any old A-level textbook).
But much better is this way. Let S=the sum to infinity.
S = 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + ...
then
S/2 = 1/4 + 1/8 + 2/16 + 3/32 + 5/64 + 8/128 + ...
then get 2S from the first line:
2S = 1 + 1/2 + 2/4 + 3/8 + 5/16 + 8/32 + 13/64 + ...
This is the big line:
2S - S/2 = S + 1
Done!
One of my friends said he has seen the solution in the book
Fibonacci and Lucas numbers, and the Golden Section: Theory and Applications by S Vajda (very old book).
In general..
1/r + 1/r^2 + 2/r^3 + 3/r^4 + 5/r^5 + ....
= r/[(r^2) - (r+1)]
Anyone...
(1*1)/2 + (2*1)/4 + (3*2)/8 + (4*3)/16 + ..... = ???
i think yr friend got the general solution from generalising one of the method above.
i can do the other series, but i'm at college at the moment. wait till i get home...
give us a clue...
I myself, only know the answer...I still don't know.. how it is derived or a proof for it.(actually I forgot about the whole thing...will restart my brain tomorrow)
May I help you with sigma notation will do. I'm not as elderly as most of you are, please take my advice:
Code:infinity sign
e (1/2)^n=(1/2)+(1/4)......
n=1