How do you test the convergence for sigma (n^2 +1)^n over n^(2n)?

Question

anonymous · Answer

(1+1/n^2)^n

anonymous · Answer

as n will increase, the value will be nearer and nearer to 1

anonymous · Answer

What test did you use? test for divergence, AST, etc? i have to use a test

anonymous · Answer

I haven't done this kind of stuff for a while, but I believe that I would have used L'Hopital's rule to test for convergence, since both the bottom and top go to infinity as n gets infinitely larger.

anonymous · Answer

@xingtheking: oh so youre using test for divergence right?

anonymous · Answer

I believe so; just looking at a site now about the so-named 'test for divergence'; at a basic level, the test just says that if the limit of your sum's equation (at infinity) is not 0 (or doesn't exist) then your sum will diverge.

anonymous · Answer

On the other hand, the logical opposite, that 
-'if the limit of your equation is zero, then the equation converges'
is NOT necessarily true, as in the case of 
$$\sum1/n$$

anonymous · Answer

Oh because 1/n is harmonic p series thats right