# Convergence of infinite product

1. Mar 4, 2007

### AKG

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 $\sum _{n=1} ^{\infty} \log (n \sin (1/n))$ converges
- iff the sum $\sum _{n=1} ^{\infty}| \log (n \sin (1/n))|$ converges
- iff the sum $\sum _{n=1} ^{\infty}|n \sin (1/n) - 1|$ converges

I can't figure out what to compare any of these to to show they converge. Any hints?

2. Mar 4, 2007

### StatusX

I would use the last one with a taylor expansion for sin.

3. Mar 4, 2007

### AKG

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:

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

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

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

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

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

$$= \sum _{n=1} ^{\infty} \sum _{m=2} ^{\infty} \frac{1}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}$$

$$= \sum _{m=2} ^{\infty} \sum _{n=1} ^{\infty} \frac{1}{(2m-1)!}\left ( \frac{1}{n}\right )^{2m-2}$$

$$= \sum _{m=2} ^{\infty}\left ( \frac{1}{(2m-1)!} \sum _{n=1} ^{\infty} n^{-(2m-2)} \right )$$

$$= \sum _{m=2} ^{\infty} \frac{\zeta (2m-2)}{(2m-1)!}$$

$$\leq \sum _{m=2} ^{\infty} \frac{\zeta (2m-2)}{(m-1)!}$$

$$= \sum _{m=1} ^{\infty} \frac{\zeta (2m)}{m!}$$

$$\approx 2.407447$$

where $\zeta$ is the Riemann $\zeta$ function, and the last approximation is given by Mathworld.

4. Mar 4, 2007

### StatusX

You're over complicating it. They taylor series can be truncated to get:

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

just use the limit comparison test.

5. Mar 5, 2007

### tehno

Remember that :
$$nsin\frac{1}{n}\leq\lim_{n\to \infty}nsin\frac{1}{n}=1$$

So we have straightforwardly $\prod _{n=1} ^{\infty} n\sin (1/n)\leq 1$

6. Mar 5, 2007

### AKG

Thanks, that worked out perfectly.

7. Mar 5, 2007

### AKG

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. Mar 6, 2007

### tehno

?

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

$$\left(1\cdot sin\frac{1}{1}\right)\left(2\cdot sin\frac{1}{2}\right)\left(3\cdot sin\frac{1}{3}\right)...$$

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: Mar 6, 2007
9. Mar 6, 2007

### AKG

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. Mar 6, 2007

### Dick

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. Mar 6, 2007

### StatusX

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. Mar 6, 2007

### Dick

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. Mar 6, 2007

### StatusX

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. Mar 6, 2007

### Dick

That's a good point. But then the product of 1/n doesn't converge either. Too bad, I guess.

15. Mar 7, 2007

### tehno

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:

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

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

Therefore we have:

$$\prod \frac{sin(1/n)}{\frac{1}{n}}>\prod \frac {sin(1/n)}{tg(1/n)}=\prod cos \frac{1}{n}$$

AKG can you finish the work and show now that

$$\sum_{n=1}^{\infty}ln\ cos\frac{1}{n}$$

converges?

Last edited: Mar 7, 2007
16. Mar 7, 2007

### AKG

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. Mar 7, 2007

### tehno

But you can easily make it work even without that typo of mine.
Hehe.
Don't you agree that:

$$\frac{1}{n}<tg\frac{1}{n}$$

is true for $n>0;n\in\mathbb{N}$ ?

18. Mar 8, 2007

### zoki85

True!
The only correction to be made relates to the interval "typo".

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

$$-\sum_{n=1}^{\infty} ln(cos\frac{1}{n})$$
This sum converges by Cauchy's criterion (demonstrable in one line) and the proof is complete.
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"