21Firstly Life is not a race and the 2nd one is \sum_{n=1}^{inf}{}\frac{1+2+...(2n-1)}{1^{3}+ 2^{3}+ ...+n^{3}} =\sum_{n=1}^{inf}{}\frac{4}{(n+1)^{2}} = 4(\xi (2)-1) = ??
The first one perhaps requires definite integrals.
8 Answers
21
62yup it is and that is what that symbol used stands for :P
reimann zeta function...
62for the 1st one, a useful in the exam method is
\frac{{({(2n)}^n)}^{1/n}}{n}\le f(n)\le\frac{{({(3n)}^n)}^{1/n}}{n}
\\\2\le f(n)\le3
so the answer is clearly e...
But the better approach will be to take log and apply sum as a limit...convert to an integral and then solve ....
262k... found my mistake.....
@nishant - the exam. method was helpful...
30@ Nishant Sir - But then how do we find the limiting value of f(n), because at ∞ limit will lie between 2 and 3 right?
262just verify with the given options... answer comes out to be B (>2 and <3)