# Prove that $\lim_{n\to \infty}{\left(1+\frac{1}{n^2}\right)^{n^3}}=+\infty$

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.

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

Using my comment: for almost all $\;n\in\Bbb N\;$ , we have that (since $\;e=2.7...\;$)
$$2.5\le\left(1+\frac1{n^2}\right)^{n^2}\le3\implies\left[\left(1+\frac1{n^2}\right)^{n^2}\right]^n\ge(2.5)^n\xrightarrow[n\to\infty]{}\infty$$
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 $n \to \infty$.
$\left(1+\frac{1}{n^2}\right)^{n^3} \ge 1+n$.