I got 2/3. take out the 1/9 and you get Sn as 1/9(1/4 x2/5x3/6...)
the numerators and denominators of this cancel except for the first 1x2x3 and hence
sn=1/9x6=2/3
Let
then find
Mahan Ghobadi
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I got 2/3. take out the 1/9 and you get Sn as 1/9(1/4 x2/5x3/6...)
the numerators and denominators of this cancel except for the first 1x2x3 and hence
sn=1/9x6=2/3
Unfortunately, the answer is not .
You have an interesting idea but when it gets to infinite product or sum, weird things happen. I believe, you can't in general use telescoping method for infinite sums or products.
Consider this example:
let
(By Leibniz test s is finite, )
but dividing s by 2 we get the series, t:
by REARRANGING the terms we get :
which is a contradiction. it turns out rearranging the terms in an infinite series requires more care than when we deal with finite series.
So I suggest, you get the general form of then use limit to find the answer.
Hint: use the same cancellation pattern but for finite n.
Last edited by Mahan1; 20 Feb 2017 at 12:43 AM.
Mahan Ghobadi
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It will depend on whether the infinite sum converges or not.
If it diverges, then possibly you can rearrange it to make it converge to whatever value you want, including contradictions in some cases.
In this case, a quick test for a infinite series whether it converges, is does the nth term tend towards 0.
(clearly if it is an alternating series, there is a different test)
Last edited by dan964; 20 Feb 2017 at 8:07 AM.
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Here's my thinking on it. I got the same answer as Kingom but I haven't done many infinite product questions...
Last edited by BenHowe; 20 Feb 2017 at 4:05 PM.
1st Year BAppFinBActStud @ MQ
Mahan Ghobadi
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An interesting extension is:
Last edited by seanieg89; 20 Feb 2017 at 1:40 PM.
They are the same, but when you found the infinite product with your method, you changed around the order of multiplication, and for infinite products, this can change the answer (like how it can for infinite sums (see: https://en.wikipedia.org/wiki/Riemann_series_theorem); note that an infinite product of positive terms will converge to a non-zero number if and only if the infinite sum of logs of terms converges to a real number (and "diverges" to 0 if the infinite sum of logs of terms diverges to negative infinity). You can find more information at the Wikipedia page for infinite products: https://en.wikipedia.org/wiki/Infinite_product .).
And don't worry, infinite products aren't really in the HSC syllabus.
Last edited by InteGrand; 20 Feb 2017 at 5:12 PM.
It is always zero when the degree of the denominator exceeds the degree of the numerator by exactly one. (As in the original question in this thread, although to prove this more general statement you do not have telescoping at your disposal.)
Your second product does not satisfy P(n) < Q(n) for all n, which is a hypothesis of the question. (If one of the factors is zero, then certainly the infinite product tends to zero.)
Last edited by seanieg89; 20 Feb 2017 at 10:34 PM.
By the degree condition nP(n)/Q(n) tends to some positive constant limit (the ratio of leading coefficients.) Let's call this 2c.
Then for sufficiently large n (say n > K), we have nP(n)/Q(n) > c, and (1-c/n) > 0.
Then for N > k:
It's a common trick to convert infinite products into infinite sums which are in some ways nicer to work with.
There were also two facts used that I didn't bother writing proofs of (you should try to make sure you understand why they are true):
1. log(1+x) =< x for all x > -1 where these things are defined.
This is an MX2 level application of calculus.
2. The harmonic series (1+1/2+1/3+...) diverges.
This can be done in many many different ways. If you like calculus, then you can bound the partial sums of this series below by an integral involving 1/x. The resulting quantity will grow like log(x) and hence diverge to +infinity.
Last edited by Mahan1; 21 Feb 2017 at 1:04 PM.
Mahan Ghobadi
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The problem is the for infinite products doesn't hold.
same goes for infinite sums.
A similar case is swapping sums and integrals, which requires the following theorem called the Dominated Convergence Theorem...
https://en.wikipedia.org/wiki/Domina...rgence_theorem
(sorry was lazy and linked Wikipedia)
and refer to the answer to this question:
http://math.stackexchange.com/questi...e-interchanged
In terms of what Mahan is alluding to in his above post, is defining an infinite series or product, in terms of its "partial sums" or partial products. We can consider these partial sums/partial products in turn, as a infinite sequence.
Now infinite sequences are fairly easy to grapple with once we understand function and limits and that a sequence
is a function.
Note the codomain (place where we want the function to end up in), of which the range is a part of can be or (although it does become a bit interesting when considering limits in the later), or some portion of either e.g. the rational numbers.
And hence then we can take a limit on the function, that is the sequence of partial sums/products as ....
Last edited by dan964; 26 Feb 2017 at 10:08 PM.
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2014 HSC
2015 Gap Year
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Links:
View English Notes (no download)
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P_N is always positive by definition, so we just need to bound it above by something that tends to zero.
In the notation I introduced earlier, this upper bound is:
C/exp(h_k-h_N)
where h_n = 1 + 1/2 + ... + 1/n denotes the n-th partial sum of the harmonic series, and C and k are constants.
(k and K mean the same thing, I changed my mind about capitalisation and couldn't be bothered to fix all of them.)
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