1. ## convergence of sequence

(Xn) = (1 + 1/n^2)^n^2

i) does it converge?
ii) find the limit.

attempt at a solution for i):
- prove that it is monotonic and bounded
=> convergence
- bounded:
|(Xn)| <= M , all n in Naturals
(1 + 1/n^2)^n^2 <= M
n^2ln((n^2 + 1)/n^2) <= lnM
n^2ln(n^2 + 1)-n^2ln(n^2) <= lnM
n^2ln(n^2 + 1)-n^2ln(n^2) <= n^2ln(n^2 + 1) so,
n^2ln(n^2 + 1) <= lnM
not sure how to continue...

ii) have yet to try

thanks

2. Originally Posted by sprinks13
(Xn) = (1 + 1/n^2)^n^2

i) does it converge?
ii) find the limit.

attempt at a solution for i):
- prove that it is monotonic and bounded
=> convergence
- bounded:
|(Xn)| <= M , all n in Naturals
(1 + 1/n^2)^n^2 <= M
n^2ln((n^2 + 1)/n^2) <= lnM
n^2ln(n^2 + 1)-n^2ln(n^2) <= lnM
n^2ln(n^2 + 1)-n^2ln(n^2) <= n^2ln(n^2 + 1) so,
n^2ln(n^2 + 1) <= lnM
not sure how to continue...

ii) have yet to try

thanks
By choosing $\displaystyle n^2$ terms of the sequence $\displaystyle \left( 1 + \frac{1}{n} \right)^n$ we form the subsequence $\displaystyle \left( 1 + \frac{1}{n^2} \right)^{n^2}$. Since $\displaystyle \left( 1 + \frac{1}{n} \right)^n \to e$ it means each and every subsequence converges to the same limit.