Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$ 
How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$
 A: By Bernoulli's inequality
$$\left(1+\dfrac{1}{n}\right)^{n^2} > 1+n^2\frac{1}{n}=1+n$$
A: $$ \log( (1 + 1/n)^n ) \to \log(e) = 1$$ So 
$$ n^2 \log( 1 + 1/n) = n \log((1 + 1/n)^n) \to \infty$$
A: Hint: Use Taylor series of  $\ln(1+x)$ at the point $x=0$ and see what you get. Here is the Taylor series
$$ \ln(1+x) = x-{\frac {1}{2}}{x}^{2}+O \left( {x}^{3} \right) . $$
I think it is clear now.
Added:
$$\ln( 1+\frac{1}{n} )= \frac{1}{n}-{\frac {1}{2}}{\frac{1}{n^2}}+{\frac {1}{2}}{\frac{1}{n^2}}-\dots$$
$$ \implies  n^2\ln( 1+\frac{1}{n} )= {n}-{\frac {1}{2}}{}+{\frac {1}{3}}{\frac{1}{n}}-\dots$$
$$ \implies \lim_{n\to \infty} n^2\ln( 1+\frac{1}{n} ) = \infty .$$
A: $$n^{2}\log \left( 1+\frac{1}{n}\right) \sim n.$$
A: Another approach. Since
$$\left(1+\frac1n\right)^n\xrightarrow [n\to\infty]{}e>2.5$$
there exists $\,M\in\Bbb N\,$ s.t.
$$n>M\;\implies\; \left(1+\frac1n\right)^n>2$$
and thus, for $\,n>M$ :
$$\left(1+\frac1n\right)^{n^2}=\left[\left(1+\frac1n\right)^n\right]^n>2^n\xrightarrow [n\to\infty]{}\infty$$
A: $$\log\left(1+\frac1n\right)=-\log\left(1-\frac1{n+1}\right)\geqslant\frac1{n+1}$$
A: $n^2\log\left(1+\dfrac{1}{n}\right)$ tend to " $\infty \cdot 0$" as $n \rightarrow \infty$.
$\frac{\log\left(1+\dfrac{1}{n}\right)}{1/n^2}$ tend to "$\frac{0}{0}$" as $n \rightarrow \infty$.
By L'hopital's rule, the limit is equivalent to the limit of,
$\frac{\left( \frac{-1/n^2}{1+\frac{1}{n}} \right)}{\frac{-2}{n^3}} = \left( \frac{n}{2} \right) \left( \frac{1}{1+\frac{1}{n}} \right)$, which is "$\infty \cdot 1$" $= \infty$.
