Finding limit of S (1 Viewer)

Mahan1 · Feb 19, 2017

Let
$S_{n} = \frac{5}{9} \ \frac{14}{20} \ \frac{27}{35} \ ... \ \frac{2n^{2}-n-1}{2n^{2}+n-1}$

then find
$\lim_{n\to\infty}S_{n}$

Kingom · Feb 19, 2017

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

Mahan1 · Feb 20, 2017

Kingom said:
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 $\frac{2}{3}$ .
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
$s = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$ (By Leibniz test s is finite, $s\neq 0$ )

but dividing s by 2 we get the series, t:

$t = \frac{1}{2}- \frac{1}{4}+\frac{1}{6}+...$

$t= (1-\frac{1}{2}) -\frac{1}{4} +(\frac{1}{3}- \frac{1}{6}) +...$
by REARRANGING the terms we get :

$t = 1 -\frac{1}{2}+\frac{1}{3}- .... = s$

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 $S_{n}$ then use limit to find the answer.
Hint: use the same cancellation pattern but for finite n.

dan964 · Feb 20, 2017

Mahan1 said:
Unfortunately, the answer is not $\frac{2}{3}$ .
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
$s = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$ (By Leibniz test s is finite, $s\neq 0$ )

but dividing s by 2 we get the series, t:

$t = \frac{1}{2}- \frac{1}{4}+\frac{1}{6}+...$

$t= (1-\frac{1}{2}) -\frac{1}{4} +(\frac{1}{3}- \frac{1}{6}) +...$
by REARRANGING the terms we get :

$t = 1 -\frac{1}{2}+\frac{1}{3}- .... = s$

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 $S_{n}$ then use limit to find the answer.
Hint: use the same cancellation pattern but for finite n.

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)

BenHowe · Feb 20, 2017

Here's my thinking on it. I got the same answer as Kingom but I haven't done many infinite product questions...

$S_n=\frac{5}{9}\frac{14}{20}\frac{27}{35}...\frac{2n^{2}-n-1}{2n^{2}+n-1}\\\lim_{n\to\infty}S_n=\prod\limits_{n=2}^\infty\frac{2n^{2}-n-1}{2n^{2}+n-1}\\=\prod\limits_{n=2}^\infty\frac{(n-1)(2n+1)}{(n+1)(2n-1)}\\=(\prod\limits_{n=2}^\infty\frac{n-1}{n+1})(\prod\limits_{n=2}^\infty\frac{2n+1}{2n-1})\\\prod\limits_{n=2}^\infty\frac{n-1}{n+1}=\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}...=(1)(2)=2\\\prod\limits_{n=2}^\infty\frac{2n+1}{2n-1}=\frac{5}{3}\frac{7}{5}\frac{9}{7}...=\frac{1}{3}\\\prod\limits_{n=2}^\infty\frac{2n^{2}-n-1}{2n^{2}+n-1}=(2)(\frac{1}{3})=\frac{2}{3}$

Mahan1 · Feb 20, 2017

BenHowe said:
Here's my thinking on it. I got the same answer as Kingom but I haven't done many infinite product questions...

$S_n=\frac{5}{9}\frac{14}{20}\frac{27}{35}...\frac{2n^{2}-n-1}{2n^{2}+n-1}\\\lim_{n\to\infty}S_n=\prod\limits_{n=2}^\infty\frac{2n^{2}-n-1}{2n^{2}+n-1}\\\prod\limits_{n=2}^\infty\frac{(n-1)(2n+1)}{(n+1)(2n-1)}\\=(\prod\limits_{n=2}^\infty\frac{n-1}{n+1})(\prod\limits_{n=2}^\infty\frac{2n+1}{2n-1})\\\prod\limits_{n=2}^\infty\frac{n-1}{n+1}=\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}...=(1)(2)=2\\\prod\limits_{n=2}^\infty\frac{2n+1}{2n-1}=\frac{5}{3}\frac{7}{5}\frac{9}{7}...=\frac{1}{3}\\\prod\limits_{n=2}^\infty\frac{2n^{2}-n-1}{2n^{2}+n-1}=(2)(\frac{1}{3})=\frac{2}{3}$

You are making the same mistake.
instead of infinity you should find $S_{k}$ for finite k that means

$S_{k} = \prod_{n=2}^{k} (\frac{n-1}{n+1}) \ \prod_{n=2}^{k}(\frac{2n+1}{2n-1}) = (\frac{2}{k(k+1)})(\frac{2k+1}{3}) = \frac{4k+2}{3k^{2}+3k}$
then
$lim_{k\to\infty} S_{k} = lim_{k\to\infty} \frac{4k+2}{3k^{2}+3k} = 0$

seanieg89 · Feb 20, 2017

An interesting extension is:

$Find: $\\ \prod_{n=1}^\infty \left(1-\frac{P(n)}{Q(n)}\right)\\ \\ $where $P(n)<Q(n)$ are polynomials that are positive at positive integers n, and $\deg(P)=\deg(Q)-1.$\\ \\ What can you say about the limit if $\deg(Q)>\deg(P)+1$?$

BenHowe · Feb 20, 2017

Mahan1 said:
You are making the same mistake.
instead of infinity you should find $S_{k}$ for finite k that means

$S_{k} = \prod_{n=2}^{k} (\frac{n-1}{n+1}) \ \prod_{n=2}^{k}(\frac{2n+1}{2n-1}) = (\frac{2}{k(k+1)})(\frac{2k+1}{3}) = \frac{4k+2}{3k^{2}+3k}$
then
$lim_{k\to\infty} S_{k} = lim_{k\to\infty} \frac{4k+2}{3k^{2}+3k} = 0$

So k is just any random number that satisfies $2\leq{k}<\infty$ ? And if so you need to do this because it doesn't make sense for k to be infinity...? Don't quite understand what you mean

InteGrand · Feb 20, 2017

BenHowe said:
So k is just any random number that satisfies $2\leq{k}<\infty$ ?

$\noindent Basically. He's finding an expression for the partial product up to a finite number $k$ (obviously an integer) and then taking the limit of this as $k\to \infty$, as the original question asks (and other methods can fail as we have seen above).$

BenHowe · Feb 20, 2017

seanieg89 said:
An interesting extension is:

$Find: $\\ \prod_{n=1}^\infty \left(1-\frac{P(n)}{Q(n)}\right)\\ \\ $where $P(n)<Q(n)$ are polynomials that are positive at positive integers n, and $\deg(P)=\deg(Q)-1.$\\ \\ What can you say about the limit if $\deg(Q)>\deg(P)+1$?$

Is it always $0$ ? I don't know how to explain it for all cases but I just tried $\prod\limits_{n=1}^{\infty}1-\frac{1}{n^{2}}$ because I think it satisfies the above condition and I got $\infty\times0$ , so I thought it'd be $0$ ?

BenHowe · Feb 20, 2017

InteGrand said:
$\noindent Basically. He's finding an expression for the partial product up to a finite number $k$ (obviously an integer) and then taking the limit of this as $k\to \infty$, as the original question asks (and other methods can fail as we have seen above).$

But I don't quite understand the difference between $\prod\limits_{n=1}^{\infty}\frac{2n^{2}-n-1}{2n^{2}+n-1}$ and $\lim_{x\to\infty}S_n$ .

InteGrand · Feb 20, 2017

BenHowe said:
But I don't quite understand the difference between $\prod\limits_{n=1}^{\infty}\frac{2n^{2}-n-1}{2n^{2}+n-1}$ and $\lim_{x\to\infty}S_n$ . Sorry if this seems stupid my teacher hadn't taught ext1/2 maths before so I did my best, but I have some gaps.

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.

seanieg89 · Feb 20, 2017

BenHowe said:
Is it always $0$ ? I don't know how to explain it for all cases but I just tried $\prod\limits_{n=1}^{\infty}1-\frac{1}{n^{2}}$ because I think it satisfies the above condition and I got $\infty\times0$ , so I thought it'd be $0$ ?

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.)

BenHowe · Feb 21, 2017

seanieg89 said:
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.)

Sorry to bother you, would you be able to post a proof then?

BenHowe · Feb 21, 2017

InteGrand said:
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.

So what you're telling me is that $\prod\limits_{n=1}^{\infty}\frac{2n^{2}-n-1}{2n^{2}+n-1}\neq(\prod\limits_{n=1}^{\infty}\frac{n-1}{n+1})(\prod\limits_{n=1}^{\infty}\frac{2n+1}{2n-1})$ ? Is there a rule kinda like a product of matrices going on here?

seanieg89 · Feb 21, 2017

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:

$P_N=\prod_{j=1}^K (1-\frac{P(j)}{Q(j)})\cdot \prod_{j=k+1}^N (1-\frac{P(j)}{Q(j)})\\ \\ =C \prod_{j=k+1}^N (1-\frac{P(j)}{Q(j)})\\ \\ \Rightarrow \log(P_N)\leq \log(C)+\sum_{j=k+1}^N \log(1-\frac{c}{j})\\ \\ \leq \log(C)-\sum_{j=k+1}^N \frac{c}{j}.\\ \\ $This quantity tends to $-\infty$ from the divergence of the harmonic series, and so by the squeeze law we can conclude that $P_N\rightarrow 0. $$

BenHowe · Feb 21, 2017

seanieg89 said:
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:

$P_N=\prod_{j=1}^K (1-\frac{P(j)}{Q(j)})\cdot \prod_{j=k+1}^N (1-\frac{P(j)}{Q(j)})\\ \\ =C \prod_{j=k+1}^N (1-\frac{P(j)}{Q(j)})\\ \\ \Rightarrow \log(P_N)\leq \log(C)+\sum_{j=k+1}^N \log(1-\frac{c}{j})\\ \\ \leq \log(C)-\sum_{j=k+1}^N \frac{c}{j}.\\ \\ $This quantity tends to $-\infty$ from the divergence of the harmonic series, and so by the squeeze law we can conclude that $P_N\rightarrow 0. $$

You lost me... Why did you go to logs and sigma?

seanieg89 · Feb 21, 2017

BenHowe said:
You lost me... Why did you go to logs and sigma?

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.

Mahan1 · Feb 21, 2017

BenHowe said:
So what you're telling me is that $\prod\limits_{n=1}^{\infty}\frac{2n^{2}-n-1}{2n^{2}+n-1}\neq(\prod\limits_{n=1}^{\infty}\frac{n-1}{n+1})(\prod\limits_{n=1}^{\infty}\frac{2n+1}{2n-1})$ ? Is there a rule kinda like a product of matrices going on here?

$\textrm{It is one of the definition of product/sum of infinite terms:}$
$\textrm{Let me give you one example of infinite product,}$

$\frac{3}{2}\ \frac{5}{4} \ \frac{7}{6}\ \frac{9}{8}\ ....$

$\textrm{note that every term is greater than 1 so either the product is infinite or a finite number greater than 1.}$ $\textrm{if we rearrange the terms, we get }$

$(\frac{1}{2}\ \frac{1}{4} \ \frac{1}{6}\ \frac{1}{8}\ ....)(3 . 5 .7 .9 ...)$

$\frac{1}{2}\ (\frac{1}{2} \ \frac{1}{3}\ \frac{1}{4}\ ....)(3 .5 .7 . 9 ...)$

$\frac{1}{2}\ (\frac{1}{2} \ \frac{1}{4}\ \frac{1}{6}....) = 0$

$\textrm{As you can see we found contradictory answers,}$ $\textrm{so to avoid such problems we have to define infinite sum and product.}$

$\textrm{We do know how to multiply finite numbers but we do not know how to multiply infinitely many terms.}$ $\textrm{So as a definition we define it as}$

$\prod_{n=1}^{\infty}a_{1} = lim_{n\to\infty}S_{n} \ \ \textrm{where} S_{n} = a_{1}..a_{n}$

$\textrm{this definition intuitively makes sense and is practical}$ $\textrm{ since we know how to do a finite product and know how to take a limit.}$

$\textrm{Look at the geometric sum, we first find the sum of the first n terms}$ $\textrm{ then we take the limit of n as n goes to infinity, in order to find the infinite sum.}$

$\textrm{just look at this approach as one way of making sense of infinite sum/product.}$

$\textrm{There are other ways that are less intuitive and give us some surprising results such as }$

$1+2+3+4+... = -\frac{1}{12} \ \textrm{or} \ \ 1-1+1-1+1-1+... = \frac{1}{2}$

dan964 · Feb 22, 2017

Mahan1 said:
You are making the same mistake.
instead of infinity you should find $S_{k}$ for finite k that means

$S_{k} = \prod_{n=2}^{k} (\frac{n-1}{n+1}) \ \prod_{n=2}^{k}(\frac{2n+1}{2n-1}) = (\frac{2}{k(k+1)})(\frac{2k+1}{3}) = \frac{4k+2}{3k^{2}+3k}$
then
$lim_{k\to\infty} S_{k} = lim_{k\to\infty} \frac{4k+2}{3k^{2}+3k} = 0$

The problem is the $\prod AB = \prod A \prod B$ 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/Dominated_convergence_theorem
(sorry was lazy and linked Wikipedia)
and refer to the answer to this question:
http://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-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. $\{u_{n}\}: \mathbb{N} \mapsto \mathbb{R}$
Note the codomain (place where we want the function to end up in), of which the range is a part of can be $\mathbb{R}$ or $\mathbb{C}$ (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 $n \in \mathbb{N}; n \rightarrow \infty$ ....

Finding limit of S (1 Viewer)

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