Sweet . But can you explain it in English?.... but how did he arrive a that?

Will do that in the next post.

But meanwhile, taking Ramanaji suggestion, let me comment on one topic of math,

continued fraction, which does not get too much attention these days. I learned this in my high school (or earlier) like all others in those days, but the subject is not taught here in US and I don't think people do that in India now a days either.

This is why when I heard a very similar problem (to sigma(n)) I knew the answer almost right away .. (I'll come back to that at the end of the post).

The theory was, according to wiki and other sources, developed by Bhaskara II. People in India learned it, others did not (or not that much)

According to wiki:

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan After he [Hardy] saw Ramanujan's theorems on continued fractions on the last page of the manuscripts, Hardy commented that the "[theorems] defeated me completely; I had never seen anything in the least like them {People in Cambridge were unfamiliar with that } before".[58] He figured that Ramanujan's theorems "must be true, because, if they were not true, no one would have the imagination to invent them".[58] Hardy asked a colleague, J. E. Littlewood, to take a look at the papers....

Wiki (or Wolfram or such sources) has good article on continued fraction.

And using this the sigma problem becomes very easy..

sqrt(2) = 1 + 1/( 2+ 1/( 2+ 1/(2+1/(......)))).....

(see:

http://en.wikipedia.org/wiki/Square_root_of_2#Continued_fraction_representationWith first convergents: 1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408 ...

(Denominators are same as the series posted before {1,2,5,12,29,70..}

Each fraction is pretty close to sqrt(2), .. so if the fraction is (a/b)

a^2/b^2 is close to 2, in fact, difference between 2b^2 and a^2 is 1..

QED.

(PS - the logic of the relationship given by Brahamgupta (a_n = 2*a_(n-1)+a_(n-2)) becomes quite clear here)