Series for pi?

By Pooya Farshim (P2572) on Sunday, January 28, 2001 - 02:32 pm :

Prove or disprove:

\sum_{n = 1}^{\infty} \frac{sgn \sin n}{n} = \frac{π^{2}}{7}

By Pooya Farshim (P2572) on Monday, January 29, 2001 - 12:58 pm :

The above sum gets close to

π^{2} / 7

, but not quite close enough. How can I prove:

the sum is convergent;
its limit is not $π^{2} / 7$ ?

By Michael Doré (Md285) on Monday, January 29, 2001 - 11:08pm:

Hi Pooya,

This is quite interesting. I can't believe the fact that it gets so close to

π^{2} / 7

is a co-incidence yet the error never seems to go below about 0.1%.

To show it converges - first prove that in the sequence {sgn sin n} the 1s and -1s always occur in blocks of 3 or 4, and that between two blocks of 4 "1s" must occur a block of 4 "-1s" and vice versa. From that try and get some sort of upper and lower bound on the sum and hence show it converges.

By Pooya Farshim (P2572) on Tuesday, January 30, 2001 - 01:31 pm :

Thanks Michael. I came up with this series while playing with Riemann's re-arrangement theorem. I found it quite interesting that it goes so close to

π^{2} / 7

. But how can I prove that the limit is not this value (I calculated the sum to

n = 10^{7}

and it's not as close as to

π^{2} / 7

as it should be.)

By Pooya Farshim (P2572) on Tuesday, January 30, 2001 - 04:44 pm :

Prove or disprove:

\int_{1}^{\infty} \frac{sgn \sin x}{x} dx = \ln 2

By Pooya Farshim (P2572) on Tuesday, January 30, 2001 - 05:38 pm :

Well Michael! How about this one? If

J

is the value of the above integral from 1 to

n

, then

e^{J}

is a rational whenever

sgn \sin n

is positive and is a rational multiple of

π^{2}

whenever

sgn \sin n

is negative (clearly!) Is there a connection between this

π^{2}

and the

π^{2}

we had above?

By Pooya Farshim (P2572) on Tuesday, January 30, 2001 - 05:54 pm :

By the way the value of the sum to n=10⁸ is 1.409238392475385

By Pooya Farshim (P2572) on Tuesday, January 30, 2001 - 06:03 pm :

The convergence is obvious! The value of the sum is always between sum to 3 and sum to 6. (i.e. between 55/30 and 73/60)

By Kerwin Hui (Kwkh2) on Tuesday, January 30, 2001 - 06:41 pm :

In the following argument, log is taken base e.

$\int_{1}^{\infty} (sign \sin x) / x dx$

$= \int_{1}^{π} (1 / x) dx - \int_{π}^{2 π} (1 / x) dx + \int_{2 π}^{3 π} - \dots$

$= \log π - \log 2 + \log 3 / 2 - \log 4 / 3 + \dots$

$= \log π - \log (2 * (2 / 3) * (4 / 3) * (4 / 5) * \dots)$

$= \log π - \log (Π_{n = 1}^{\infty} (4 n^{2} / 2 n^{2} - 1))$

$= \log π - \log (Π_{n = 1}^{\infty} 1 / (1 - 1 / (4 n^{2})))$

From Euler's sine product [ $\sin x = x Π_{n = 1}^{\infty} (1 - x^{2} / (n^{2} π^{2}))$ ], we have that when $x = 1 / 2 π, 1 = 1 / 2 π Π_{n = 1}^{\infty} (1 - 1 / (4 n^{2}))$ , i.e.

$\int_{1}^{\infty} (sign \sin x) / x dx = \log π + \log (2 / π) = \log 2$

Kerwin

By Michael Doré (Md285) on Tuesday, January 30, 2001 - 11:24pm:

Yep, the key to that one is Euler's product, and I thought at first that your first series was also a consequence of this. I'm still not entirely convinced that your first series doesn't converge to $π^{2} / 7$ - perhaps it just converges incredibly slowly after a quick start. Or maybe there is some subtlely different series which only differs very slightly from yours which does converge to $π^{2} / 7$ .
The second identity is much easier to analyse than the first one because with the first one you need to evaluate the function at discrete points; but in the second one you can just wait till sin x changes sign and split up the integral.

I'm not quite sure what you mean by e^J is rational, where J is the integral from 1 to n. Is n an integer here?

By Pooya Farshim (P2572) on Wednesday, January 31, 2001 - 01:17 pm :

Thanks Kerwin. That was pretty good. (As Expected!)

Yes! n is an integer here.

I was thinking whether

γ

and ln(2) could play a role in the sum...

By Pooya Farshim (P2572) on Wednesday, January 31, 2001 - 01:31 pm :

The value of the sum is not

π^{2} / 7

As $sgn \sin (10^{8} + 1) = sgn \sin (10^{8} + 5) = 1$ and $sgn \sin (10^{8} + i) = - 1$ for $i = 2$ , 3, 4, the value of the sum lies between sum to $n = 10^{8} + 1$ and sum to $10^{8} + 4$ . However $π^{2} / 7$ is less than the latter.

By Brad Rodgers (P1930) on Friday, February 2, 2001 - 08:06 pm :

Could someone explain what

Π

denotes, and why that Euler's relationship holds?

Thanks,

Brad

By Michael Doré (Md285) on Friday, February 2, 2001 - 08:11pm:

$Π_{i = m}^{n} f (i) = f (m) f (m + 1) \dots f (n - 1) f (n)$

Basically it is exactly the same as $\sum$ except it is a product not a sum.

I can't offer a good explanation for why Euler's sine product holds I'm afraid - I've been trying to prove it for a while. It is a plausible relationship because the product has roots at $n π$ where $n$ is an integer, as does $\sin x$ . This was how Euler first speculated that the relationship might hold - I don't know if he proved it rigorously. I'll keep trying and get back to you if I can prove it.

By Pooya Farshim (P2572) on Friday, February 2, 2001 - 11:36 pm :

There is a rigorous proof of the identity in one of my analysis books. It's long, complicated and refers back to some previous theorems. I'd rather accept it for the moment....

By Brad Rodgers (P1930) on Saturday, February 3, 2001 - 04:39 am :

Perhaps it's just ignorance acting at the moment, but how do we know that $2 \times (2 / 3) \times (4 / 3) \dots = Π_{1}^{\infty} 4 n^{2} / (4 n^{2} - 1)$ ?

Brad

By Kerwin Hui (Kwkh2) on Saturday, February 3, 2001 - 06:24 pm :

Brad,

Grouping the terms into blocks of 2, and a bit of arithmetic should convince you of why it is true.

Kerwin

By Kerwin Hui (Kwkh2) on Saturday, February 3, 2001 - 06:44 pm :

The proof of Euler's sine product requires contour integration and various theorems concerning complex analysis. which is probably in 2nd year university.