how do u find the value of 'sigma' from n=-infinity to infinity for (1/2)^(2 mod n)..
sorry wasn't able to put it in the right format..

Is that it?


\sum^n _{-\infty\to\infty} \(\({\frac12}\)^{2|n|}\)

You can try the LaTeX (http://www.forkosh.com/mimetextutorial.html) tutorial...

The most logical interpretation I can come up with for (2 mod n) is 2 modulo n, which makes no sense whatsoever. Unless the sum is supposed to diverge, then it's prefectly logical. Because for |n|>to the expression (2 mod n) is simply 2, meaning the sum is the same as (and excuse me for being too lazy to use LaTeX):
[the sum from n=-inf to -3 for (1/4)] + [the sum from n=-2 to 2 for (1/2)^(2 mod n)] + [the sum from n=3 to inf for (1/4)]

The first and past part of that expression diverge to infinity.

If |n| were an absolute value, then we could write:

2\sum_{0}^{\infty}\frac{1}{4^{|n|}}=\frac{8}{3}

What we have is a geometric series. \frac{1}{1-\frac{1}{4}}=\frac{4}{3}

Just a thought. I am uncertain of the 'mod' as well.

The notation given by jerry is exactly right.. The answer given is 5/3... I don't know how to arrive at it..

The notation given by jerry is exactly right..The answer given is 5/3...i dont know how to arrive at it..

Galactus almost had it! I'm guessing he sees his mistake now that you say the right answer.

In doubling the sum from 0 to infinity (once to account for the negative values of n and once for the positive), he inadvertently counted n=0 twice. Hence, the real answer is actually \frac{8}{3}-1=\frac{5}{3}

Good catch, JCaron. Yes, it was an oversight.