<s>[INST] How can I show that $\lim_{n\to\infty} 2^n \left( \frac{n}{n+1} \right ) ^{n^2} = 0$? How to calculate limit of the following expression:
$$2^n \left( \frac{n}{n+1} \right ) ^{n^2} $$
I know that limit of this sequence is equal to zero, but how to show that?
 [/INST] $(\frac{n}{n+1})^{n^{2}} = ( 1 -\frac{1}{n+1})^{n^{2}}$. For sufficiently large $n$,
$(1 - \frac{1}{n+1})^{n+1}$ is very close to $e^{-1}$, so is less than $\frac{2}{5}$ as $e >2.5.$ Then $( 1 -\frac{1}{n+1})^{n^{2}} < ( 1 -\frac{1}{n+1})^{n^{2}-1} < (\frac{2}{5})^{n-1}$. So $2^{n}(\frac{n}{n+1})^{n^{2}} < \frac{2^{2n-1}}{5^{n}} < 2 \times (\frac{4}{5})^{n-1}$, and the rightmost expression tends to $0$ as $n \to \infty.$ 
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