Convergence of infinite product

In summary, the product \prod _{n=1} ^{\infty} (n\sin (1/n)) converges and is equivalent to the limit \lim_{x\to 0}\frac{sin\ x}{x}=1. This can be shown by using the fact that the partial products are non-increasing and bounded below by zero. Further analysis can be done by comparing the product to \sum _{n=1} ^{\infty} \log (a_n), where a_n = n\sin (1/n).
  • #1

AKG

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Homework Statement



Show that [itex]\prod _{n=1} ^{\infty} (n\sin (1/n))[/itex] converges

Homework Equations



[itex]\prod _{n=1} ^{\infty}a_n[/itex] converges iff the sequence of partial products converges to a non-zero limit. Such a product converges iff [itex]\sum _{n=1} ^{\infty} \log (a_n)[/itex] converges. [itex]\sum _{n=1} ^{\infty} |log(a_n)|[/itex] converges iff [itex] \sum _{n=1} ^{\infty}|a_n-1|[/itex] converges.

The Attempt at a Solution



Since log(nsin(1/n)) is negative for all n, the product we're interested in converges:

- iff the sequence of partial products converges to a non-zero finite number
- iff the sum [itex]\sum _{n=1} ^{\infty} \log (n \sin (1/n))[/itex] converges
- iff the sum [itex]\sum _{n=1} ^{\infty}| \log (n \sin (1/n))|[/itex] converges
- iff the sum [itex]\sum _{n=1} ^{\infty}|n \sin (1/n) - 1|[/itex] converges

I can't figure out what to compare any of these to to show they converge. Any hints?
 
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  • #2
I would use the last one with a taylor expansion for sin.
 
  • #3
I had thought of that but didn't get anywhere with it. I tried again and got convergence, but it relies on facts from Mathworld that I haven't proved:

[tex]\sum _{n=1} ^{\infty}|n \sin (1/n) - 1|[/tex]

[tex] = \sum _{n=1} ^{\infty}\left | n\sum _{m=1} ^{\infty} \frac{(-1)^{m-1}}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-1}\ -\ 1\right |[/tex]

[tex] = \sum _{n=1} ^{\infty}\left | \sum _{m=1} ^{\infty} \frac{(-1)^{m-1}}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}\ -\ 1\right |[/tex]

[tex] = \sum _{n=1} ^{\infty}\left | \sum _{m=2} ^{\infty} \frac{(-1)^{m-1}}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}\right |[/tex]

[tex] \leq \sum _{n=1} ^{\infty} \sum _{m=2} ^{\infty} \left | \frac{(-1)^{m-1}}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}\right |[/tex]

[tex] = \sum _{n=1} ^{\infty} \sum _{m=2} ^{\infty} \frac{1}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}[/tex]

[tex] = \sum _{m=2} ^{\infty} \sum _{n=1} ^{\infty} \frac{1}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}[/tex]

[tex] = \sum _{m=2} ^{\infty}\left ( \frac{1}{(2m-1)!} \sum _{n=1} ^{\infty} n^{-(2m-2)} \right )[/tex]

[tex] = \sum _{m=2} ^{\infty} \frac{\zeta (2m-2)}{(2m-1)!}[/tex]

[tex] \leq \sum _{m=2} ^{\infty} \frac{\zeta (2m-2)}{(m-1)!}[/tex]

[tex] = \sum _{m=1} ^{\infty} \frac{\zeta (2m)}{m!}[/tex]

[tex] \approx 2.407447[/tex]

where [itex]\zeta[/itex] is the Riemann [itex]\zeta[/itex] function, and the last approximation is given by Mathworld.
 
  • #4
You're over complicating it. They taylor series can be truncated to get:

[tex]|n \sin \left( \frac{1}{n} \right)-1| = \frac{1}{6 n^2} + O\left(\frac{1}{n^4} \right) [/tex]

just use the limit comparison test.
 
  • #5
AKG said:

Homework Statement



Show that [itex]\prod _{n=1} ^{\infty} (n\sin (1/n))[/itex] converges.

Any hints?
Remember that :
[tex]
nsin\frac{1}{n}\leq\lim_{n\to \infty}nsin\frac{1}{n}=1[/tex]

So we have straightforwardly [itex]\prod _{n=1} ^{\infty} n\sin (1/n)\leq 1[/itex]
 
  • #6
StatusX said:
You're over complicating it. They taylor series can be truncated to get:

[tex]|n \sin \left( \frac{1}{n} \right)-1| = \frac{1}{6 n^2} + O\left(\frac{1}{n^4} \right) [/tex]

just use the limit comparison test.
Thanks, that worked out perfectly.
 
  • #7
tehno said:
Remember that :
[tex]
nsin\frac{1}{n}\leq\lim_{n\to \infty}nsin\frac{1}{n}=1[/tex]

So we have straightforwardly [itex]\prod _{n=1} ^{\infty} n\sin (1/n)\leq 1[/itex]
Recall that per the definition I gave for convergence of a product, the partial products must converge to a non-zero number. Your observation doesn't ensure this condition is met.
 
  • #8
?

For me ,your notation [itex]\prod _{n=1} ^{\infty} (n\sin (1/n))[/itex] stands for nothing else than:

[tex]\left(1\cdot sin\frac{1}{1}\right)\left(2\cdot sin\frac{1}{2}\right)\left(3\cdot sin\frac{1}{3}\right)...[/tex]

Do we agree about that?
All partial products are positive.
No term in the brackets exceeds 1,while still gradually increasing.
That means that partial products monotonicaly decrease .
Therefore ,what can be concluded about the convergence of the infinite product?
If you're interested just in answering wether it converges or not that's enough.
If you are interested in aproximating/evaluating the limit that's another thing.
 
Last edited:
  • #9
Recall that per the definition I gave for convergence of a product, the partial products must converge to a non-zero number. Your observation doesn't ensure this condition is met.
 
  • #10
AKG said:
Recall that per the definition I gave for convergence of a product, the partial products must converge to a non-zero number. Your observation doesn't ensure this condition is met.

No, I think he's ok. He's shown the partial products are non-increasing and bounded below (by zero). That's enough to prove there is a limit.
 
  • #11
That shows the product converges in the sense that the limit over larger and larger finite products exists and is finite. However, it seems AKG was asked to show the limit is non-zero.
 
  • #12
StatusX said:
That shows the product converges in the sense that the limit over larger and larger finite products exists and is finite. However, it seems AKG was asked to show the limit is non-zero.

I guess that is what it says. Is there a technical reason for making that restriction (aside from just being able to formulate convergence in terms of logs)?
 
  • #13
I'm guessing the equivalence with the log sum is one motivation. Also, if zero was allowed as a limit, any product which contains a single zero would converge, no matter what the other terms are, which seems to go against the infiniteness of the product (ie, we can usually ignore any finite subset of an infinite sum or product when talking about convergence).
 
  • #14
That's a good point. But then the product of 1/n doesn't converge either. Too bad, I guess.
 
  • #15
He wasn't crazy when he gave that definition.
There are other reasons ,beside the mentioned one,that infinite products with limit approaching 0 are called sometimes divergent.But,enough of that.
Point is he fails to recognize I gave him a fine method which demonstrates that limit is non-zero.I guess that must be a case becouse he repeats that my observation doesn't meet the criteria of the convergence (which I'm aware of it) . This is a homework help section and I don't give a complete solution served on the table.I rather like to help making people figure out themselfes..
The limit I wrote in first post is equivalent with:

[tex]\lim_{x\to 0}\frac{sin\ x}{x}=1[/tex]
In every introductury book on calculus proving that limit
is done by help of the trigonometry circle and inequalities:

[tex]sinx<x<tgx;x\in ( 0,\pi /2)[/tex]
or
[tex]sin\frac{1}{x}<\frac{1}{x}<tg\frac{1}{x};x\in ( 0,2/\pi)[/tex]

Therefore we have:

[tex]\prod \frac{sin(1/n)}{\frac{1}{n}}>\prod \frac {sin(1/n)}{tg(1/n)}=\prod cos \frac{1}{n}[/tex]

AKG can you finish the work and show now that

[tex]\sum_{n=1}^{\infty}ln\ cos\frac{1}{n}[/tex]

converges?
:smile:
 
Last edited:
  • #16
Sorry, but your hint did nothing to help me prove that the product converges to a non-zero limit. StatusX's tip did, and I've already solved the problem. By the way, 1/x < tan(1/x) for x in (0, 2/pi) is false, so your method doesn't work.
 
  • #17
But you can easily make it work even without that typo of mine.
Hehe.
Don't you agree that:

[tex]\frac{1}{n}<tg\frac{1}{n}[/tex]

is true for [itex]n>0;n\in\mathbb{N}[/itex] ?

:smile:
 
  • #18
True!
The only correction to be made relates to the interval "typo".

It should be [itex]\frac{1}{x}<tg\frac{1}{x}[/itex] for all [itex]x\in (\frac{2}{\pi},\infty>[/itex].
With that the method works perfectly.

[tex]-\sum_{n=1}^{\infty} ln(cos\frac{1}{n})[/tex]
This sum converges by Cauchy's criterion (demonstrable in one line) and the proof is complete.
AKG said:
Sorry, but your hint did nothing to help me prove that the product converges to a non-zero limit. StatusX's tip did, and I've already solved the problem.
I distinctly recall a famous mathematician said :"It is of higher importance to solve 1 problem by 2 different methods than to solve 200 problems by 1 method"
 

1. What is the definition of convergence of an infinite product?

The convergence of an infinite product refers to the property of an infinite sequence of numbers or terms that approaches a finite limit or value as the number of terms approaches infinity.

2. How is the convergence of an infinite product determined?

The convergence of an infinite product can be determined by applying a test for convergence, such as the ratio or root test, to the individual terms of the product. If the limit of these tests is less than 1, then the product is convergent.

3. What is the difference between absolute and conditional convergence of an infinite product?

Absolute convergence of an infinite product means that the product converges regardless of the order in which the terms are multiplied. On the other hand, conditional convergence means that the product only converges if the terms are multiplied in a specific order.

4. Can an infinite product converge to 0?

Yes, it is possible for an infinite product to converge to 0. This occurs when the limit of the terms of the product approaches 0 as the number of terms approaches infinity. However, it is important to note that not all infinite products that converge will converge to 0.

5. What are some real-world applications of infinite product convergence?

Infinite product convergence is commonly used in fields such as physics, engineering, and economics to model and analyze various phenomena. For example, it can be used to calculate the value of investments that accrue compound interest, or to determine the stability of physical systems.

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