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I'm trying to prove that $\lim_{n\to \infty}{\left(1+\frac{1}{n^2}\right)^{n^3}}=+\infty$ using the fact that if $b_n\ge a_n$ eventually, and $\lim_{n\to \infty}{a_n}=+\infty$, then $\lim_{n\to \infty}{b_n}=+\infty$, where $b_n:=n^n$.

I'm struggling to show that $b_n\ge a_n$ by induction. Is this a good method? If so, what would be the best way to proceed. Thank you in advance.

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    $\begingroup$ Do you know/can you prove that $$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}=e\;?$$ $\endgroup$ – DonAntonio Nov 21 '16 at 11:58
  • $\begingroup$ @DonAntonio The exponent is $n^3$, not $n^2$ $\endgroup$ – aL_eX Nov 21 '16 at 12:01
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    $\begingroup$ @aL_eX, DonAntonio gave you a hint! $\endgroup$ – Alex Silva Nov 21 '16 at 12:02
  • $\begingroup$ @DonAntonio haha, sorry $\endgroup$ – aL_eX Nov 21 '16 at 12:03
  • $\begingroup$ ;) ...${}{}{}{}{}{}{}{}{}$ $\endgroup$ – DonAntonio Nov 21 '16 at 12:04
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Using my comment: for almost all $\;n\in\Bbb N\;$ , we have that (since $\;e=2.7...\;$)

$$2.5\le\left(1+\frac1{n^2}\right)^{n^2}\le3\implies\left[\left(1+\frac1{n^2}\right)^{n^2}\right]^n\ge(2.5)^n\xrightarrow[n\to\infty]{}\infty$$

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By the Binomial Theorem, $$ \left(1 + \frac{1}{n^2}\right)^{n^3}\!\!\! = 1^{n^3} + n^3\cdot1^{n^3 - 1}\cdot\left(\frac{1}{n^2}\right) + \binom{n^3}{2}\cdot1^{n^3 - 2}\cdot\left(\frac{1}{n^2}\right)^2 + \cdots \geqslant n + 1. $$ Of course, this is a tragically weak inequality! But it does the job, because $n + 1 \to \infty$ as $n \to \infty$.

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The Bernoulli - inequality gives

$\left(1+\frac{1}{n^2}\right)^{n^3} \ge 1+n$.

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