Examine limit of sequence I have to examine limit of following sequence $\{a_n\}_{ n \ge 1}$ $a_n = \sqrt[n]{\sum_{k=1}^{n}{(2 -\frac{1}{k})^k}}$. We know that $\lim_{n \to \infty} \sqrt[n]{a} = 1$ for $a > 0$ and we know that $\sqrt[n]{\sum_{k=1}^{n}{(2 -\frac{1}{k})^k}} > 0$ but sequence $a_n$ is increasing (I cannot prove it but I checked it in wolframalpha) so we cannot use the squeeze theorem. Any hints?

I think you can get your way around this one using limited expansion.
Here I use : $ln(1-u) = -u -\frac{u^2}{2} +o(u^2)$ ; and $\exp(u) = 1+ u +o(u)$ , when $ u \rightarrow 0$
Let : $u_k = (2-\frac{1}{k})^k = 2^k*(1-\frac{1}{2k})^k = 2^k*\exp{[k*ln(1-\frac{1}{2k})]} = 2^k*\exp[-\frac{1}{2} -\frac{1}{8k} + o(\frac{1}{k})] $
=> $ u_k = \frac{2^k}{\sqrt{e}}*[1 -\frac{1}{8k} + o(\frac{1}{k})] $
Hence, you get : $u_k$ ~ $\frac{2^k}{\sqrt{e}}$
$\sum 2^k$ is a diverging series, so we can say that partial sums are equivalent : 
$ \sum_{k=1}^n v_k $ ~ $ \frac{1}{\sqrt{e}}*\sum_{k=1}^n 2^k $
$ \sum_{k=1}^n 2^k = 2*\frac{2^n -1}{2-1} $ ~ $2^{n+1}$
Hence : $ (a_n)^n = (\sum_{k=1}^n v_k) $ ~ $\frac{2^{n+1}}{\sqrt{e}}$
So : $ (a_n)^n = \frac{2^{n+1}}{\sqrt{e}} + o(2^{n+1}) $
=> $ (\frac{a_n}{2})^n = \frac{2}{\sqrt{e}} + o(1) $ => $ \frac{a_n}{2} = (\frac{2}{\sqrt{e}} + o(1) )^{\frac{1}{n}} \rightarrow 1$ ,  when $ n\rightarrow +\infty$
Hence $(a_n) \rightarrow 2$ , when $ n\rightarrow +\infty$
