I need a hint, the question is written in the picture, thank you.
Wait , what? I was sure it converged to e. The difference with the second sequence you wrote was that it goes faster than the second, but it's still bounded and and converges to e. At least that's what I thought....
$\displaystyle \begin{align*}\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^{n^2} & = \lim_{n \to \infty} \exp\left(n^2\ln\left(1+\dfrac{1}{n}\right)\right) \\ & = \exp\left( \lim_{n \to \infty} \dfrac{\ln\left(1+\dfrac{1}{n}\right) }{n^{-2}} \right) \\ & \stackrel{0}{\stackrel{=}{0}} \exp\left( \lim_{n\to \infty} \dfrac{ \left( \dfrac{ -n^{-2} }{ 1+\dfrac{1}{n} } \right) }{ -2n^{-3} } \right) \\ & = \exp\left( \lim_{n \to \infty} \dfrac{n}{2\left(1+\dfrac{1}{n}\right)} \right) \\ & = e^\infty = \infty\end{align*}$
Edit: I used L'Hospital's Rule on line 3, then for line 4, I multiplied top and bottom of the fraction by $\displaystyle -n^3$. Then, the limit as $\displaystyle n\to \infty$ of $\displaystyle \dfrac{n}{2\left(1+\dfrac{1}{n}\right)} = \infty$
It is very to show $\displaystyle {\lim _{n \to \infty }}{\left( {1 + \frac{1}{n}} \right)^{{n^2}}} = \infty $.
If $\displaystyle 1<r<e$ then $\displaystyle \exists K\in\mathbb{Z}^+[r<\left(1+\frac{1}{K}\right)^K<e]$.
The e sequence is increasing so $\displaystyle \forall n>K$ we have $\displaystyle r^n<\left( {1 + \frac{1}{n}} \right)^{{n^2}}} $
$\displaystyle {\lim _{n \to \infty }}{r^n} = \infty $