My Math Forum Infinite series : 1/2^(n^2)

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 April 22nd, 2010, 06:20 AM #1 Newbie   Joined: Apr 2010 Posts: 16 Thanks: 0 Infinite series : 1/2^(n^2) Can anyone help me with the sum of this infinite series sigma(1/2^(n^2)) ,n=0,1,2,3...I have calculated the answer to be 2 but my proof is a miserable one.It would be very useful if someone can come up with the right proof.
April 22nd, 2010, 11:06 AM   #2
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Re: Infinite series : 1/2^(n^2)

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 Originally Posted by Zeefinity Can anyone help me with the sum of this infinite series sigma(1/2^(n^2)) ,n=0,1,2,3...I have calculated the answer to be 2 but my proof is a miserable one.It would be very useful if someone can come up with the right proof.
The claim that the sum is equal to 2 is obviously wrong, since $2=\sum_{n=0}^\infty \frac1{2^n}$; and your is a subseries of this series, hence the sum must be smaller.
Another thing, which can be easily said about the sum, is that it is not a rational number, since the binary expansion is obviously not periodic.

 April 22nd, 2010, 11:08 AM #3 Senior Member   Joined: Jun 2009 Posts: 150 Thanks: 0 Re: Infinite series : 1/2^(n^2) $\sum_{n=0}^\infty \frac{1}{2^n} = 2$ and your sum is strictly less, about 1.564468, not 2. The value can be found in terms of Jacobi's Theta function, $\frac{\Theta_3 \Bigl(0,\frac{1}{2}\Bigr)+1}{2}$
 April 22nd, 2010, 10:58 PM #4 Newbie   Joined: Apr 2010 Posts: 16 Thanks: 0 Re: Infinite series : 1/2^(n^2) Can you tell me where I can find more about the theta function.I don't quite seem to understand what is given in wikipedia.

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