I need a hint, the question is written in the picture, thank you.

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- Dec 20th 2013, 11:28 AMdavidciprutNeed a hint! (Proving convergence)
I need a hint, the question is written in the picture, thank you.

- Dec 20th 2013, 11:51 AMromsekRe: Need a hint! (Proving convergence)
see what the natural log of it converges to.

- Dec 20th 2013, 11:54 AMdavidciprutRe: Need a hint! (Proving convergence)
what do you mean by ''natural log of it''? thank you.

- Dec 20th 2013, 11:59 AMromsekRe: Need a hint! (Proving convergence)
read it again

- Dec 20th 2013, 12:24 PMPlatoRe: Need a hint! (Proving convergence)
- Dec 20th 2013, 12:49 PMdavidciprutRe: Need a hint! (Proving convergence)
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....

- Dec 20th 2013, 12:58 PMSlipEternalRe: Need a hint! (Proving convergence)
$\displaystyle \begin{align*}\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^{n^2} & = \lim_{n \to \infty} \exp\left(n^2\ln\left(1+\dfrac{1}{n}\right)\right) \\ & = \exp\left( \lim_{n \to \infty} \dfrac{\ln\left(1+\dfrac{1}{n}\right) }{n^{-2}} \right) \\ & \stackrel{0}{\stackrel{=}{0}} \exp\left( \lim_{n\to \infty} \dfrac{ \left( \dfrac{ -n^{-2} }{ 1+\dfrac{1}{n} } \right) }{ -2n^{-3} } \right) \\ & = \exp\left( \lim_{n \to \infty} \dfrac{n}{2\left(1+\dfrac{1}{n}\right)} \right) \\ & = e^\infty = \infty\end{align*}$

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 $\displaystyle \dfrac{n}{2\left(1+\dfrac{1}{n}\right)} = \infty$ - Dec 20th 2013, 01:07 PMPlatoRe: Need a hint! (Proving convergence)
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 $ - Dec 20th 2013, 01:17 PMdavidciprutRe: Need a hint! (Proving convergence)
Wow, I feel so stupid right now... I got it... Thank you all