Eureka !!

This is a big one.

Remember my pi by 2 expression. Sigma for n =2 to infinity of 2^n/comb(2n,n) gave exactly pi by 2.

When I tried my rational number series with plus sign I got 2^(m-1) approximately. And the accuracy improved for higher values of n.

The expressions are

((n+1)^m/(n+2)^m+(n-1)^m/n^m)/(((n+1)/(n+2)+(n-1)/n)^m) and ((n)^m/(n+1)^m+(n-1)^m/n^m)/(((n)/(n+1)+(n-1)/n)^m). These expressions are almost 1/(2^(m-1)). And the exactness improved as n increased.

So if I mix the above two different ideas then it will be a great thing. Because of 2^n and 2^(m-1).

I shall be working on this.

(Many people may be confused with (n-1),n,(n+1) and (n+2) or (n-1),n and (n+1). There is nothing here really. 4 consecutive numbers (the way we count with our four fingers) and 3 consecutive numbers (the way we would have counted if we had three fingers))

(What I see is a brilliant pattern in them. And of course the approximation part. Which I claim to exist in 4 consecutive numbers and not in 3 consecutive numbers)

(To prove that there can be no general approximation formula for 3 consecutive numbers will be difficult for me. To me it is as difficult as Fermat's last theorem. a^n+b^n=c^n only for n = 1 and 2. Not for n>=3. But Sir Andrew Wiles proved the Fermat's last theorem. And he won so much praise and accolades from the whole world !!)

(Basel problem was posed in 1650 and first solved by Euler in 1734. The proof on impossibility of solving the general quintic equation was given by Abel and the problem was with humanity for 250 years. Fermat's last theorem was posed in 1637. And solved by Sir Andrew Wiles in 1994)

("There can be no general approximation formula for 3 consecutive numbers". Can Indian mathematicians solve this one ? )

((1(+/-)x)^n is approximately 1(+/-)nx for small values of x is damn good and involves three numbers 1,x and n. But three consecutive numbers? Interested?)

(Also whether the quintic expressions formed by ((n+1)^m/(n+2)^m+(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) ; ((n)^m/(n+1)^m+(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m; 

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) ; ((n)^m/(n+1)^m-(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m are easily solvable or not ?)(Answer - Wolfram Alpha is able to solve these type of quintic equations. I have checked.)

(So only one poser remain. Can there be a general approximation formula involving three consecutive numbers?)

This finishes.