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?
 A: $$\left(\frac{n}{n+1}\right)^{n^2}=\left(\frac{n+1}{n}\right)^{-n^2}=\left(1+\frac{1}{n}\right)^{-n^2}=\exp\left\{-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}$$
therefore
$$2^n\left(\frac{n}{n+1}\right)^{n^2}=\exp\{n\ln 2\}\cdot \exp\left\{-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}=\exp\left\{n\ln 2-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}=\exp\left\{n\underbrace{\left(\ln(2)-\underbrace{\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}}_{\to 1}\right)}_{\to \ln(2)-1<0}\right\}\longrightarrow  0 $$
A: We have:
$$\frac{1}{n}\geq \log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\geq\frac{1}{n+1},\tag{1}$$
hence:
$$ n\log 2-n^2\log\left(1+\frac{1}{n}\right)\leq n(\log 2-1)+\left(1+\frac{1}{n}\right),\tag{2}$$
and since $\log 2-1<0$, the limit of the RHS of $(2)$ as $n\to +\infty$ is $-\infty$. 
By exponentiating $(2)$, we get that our limit is zero.
A: $(\frac{n}{n+1})^{n^{2}} = ( 1 -\frac{1}{n+1})^{n^{2}}$. 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 $n \to \infty.$ 
A: $2^n (\frac{n}{n+1})^{n^2}$ = $(\frac{n2^{1/n}}{n+1})^{n^2}$
Taking log
$n^2 \log (\frac{n2^{1/n}}{n+1})=\frac{\log (\frac{n2^{1/n}}{n+1})}{1/n^2}=\frac{1/n \log(2)+\log (n/(n+1))}{1/n^2}$
Both numerator and denominator go to $0$, applying L Hospitals's rule
$\frac{-1/n^2 \log(2) +1/(n(n+1))}{-2/n^3}  = 1/2(n \log(2))-n^2/(n+1))=(n/2)(log(2)-n/(n+1))$
But
$\lim_{n \to \infty} log(2)-n/(n+1)=log(2)-1<0$
So, $n^2 \log (\frac{n2^{1/n}}{n+1}) \to -\infty$
This proves that $n^2 \log (\frac{n2^{1/n}}{n+1})\to 0$
A: $$\lim_{n\to\infty} 2^n \left( \frac{n}{n+1} \right ) ^{n^2}=\lim_{n\to\infty}\left( 2\left(1-\frac{1}{n+1}\right)^n \right )^n=\left(\lim_{n\to\infty} 2\left(1-\frac{1}{n+1}\right)^n \right )^{\lim_{n\to\infty}n}=\left(\frac{2}{e}\right)^{\lim\limits_{n\to\infty}n}=0$$
A: $$ \begin{array} {} w_n &=& 2^n ( { n \over n+1 })^{n^2} \\ \\
 \log (w_n) &=& n \log(2) - n^2 \log(1+ \frac 1n) \\ 
&=& n \log(2) - n \log((1+ \frac 1n)^n) \\ \\
\lim_{n \to \infty} \log w_n &=& n \log (2) - n \log (e) \\
&=&n (\log (2) - 1) \\
&=&  - \infty \\
\Rightarrow \\
\lim_{n \to \infty} w_n &=& \exp(-\infty) = 0
\end{array}$$
