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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?

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  • $\begingroup$ Found anything of interest in the answers below? $\endgroup$ – mvggz Nov 17 '14 at 9:52
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Firstly, for any $k_0 \in \mathbb{N}$, we have

$$a_n = \sqrt[n]{\sum_{k=1}^{n}{(2 -\dfrac{1}{k})^k}} \geq \sqrt[n]{\sum_{k=k_0}^{n}{(2 -\dfrac{1}{k_0})^k}} = \left(\dfrac{(2-1/k_0)^{k_0}(1-(2-1/k_0)^{n-k_0+1})}{1/k_0 -1}\right)^{1/n}= \left(\dfrac{(2-1/k_0)^{k_0}((2-1/k_0)^{n-k_0+1}-1)}{1-1/k_0 }\right)^{1/n} \to 2-1/k_0$$

because $$a^{1/n} \to 1, \forall a>0$$ $$\left((2-1/k_0)^{n-k_0+1}-1)\right)^{1/n} \to 2-1/k_0$$

so $\liminf a_n \geq 2-1/k_0, \forall k_0 \in \mathbb{N}$, i.e. $$\liminf a_n \geq 2$$

Secondly

$a_n = \sqrt[n]{\sum_{k=1}^{n}{(2 -\dfrac{1}{k})^k}} \leq \sqrt[n]{\sum_{k=1}^{n}{(2 -\dfrac{1}{n})^k}} = \left(\dfrac{(2-1/n)^2(1-(2-1/n)^{n})}{1/n - 1}\right)^{1/n} = \left(\dfrac{(2-1/n)^2((2-1/n)^{n}-1)}{1-1/n}\right)^{1/n} = \left(\dfrac{(2-1/n)^2}{1-1/n}\right)^{1/n}\left((2-1/n)^{n}-1\right)^{1/n}\to 2$

so $$\limsup a_n \leq 2$$

So finally $$\lim a_n =2$$

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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$

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We can apply squeeze theorem to solve it. $$\left(\lim_{n\to\infty} b_n = \lim_{n\to\infty}c_n = m \wedge \left(\exists \delta \in \mathbb{R}\right)\left(\forall n \in \mathbb{N}\right)\left( n \geq \delta \Rightarrow b_n \leq a_n \leq c_n \right)\right) \Rightarrow \lim_{n\to\infty}a_n = m$$

Now, let $b_n = \sqrt[n]{\sum_{k=1}^{n}2^k}$ and $c_n=\sqrt[n]{\sum_{k=1}^{n}\left(2^k - \frac{1}{k}\right)}$, then

$$\lim_{n\to\infty}b_n=\lim_{n\to\infty}\sqrt[n]{2^n\cdot n}=\lim_{n\to\infty}2\cdot\sqrt[n]{n}=2\\ \lim_{n\to\infty}c_n = \lim_{n\to\infty}\sqrt[n]{2^n\cdot n - \sum_{k=1}^{n}\frac{1}{k}} =\lim_{n\to\infty}2\cdot\sqrt[n]{n - \frac{H_n}{2^n}} = 2$$

$$ \left(\lim_{n\to\infty} c_n = \lim_{n\to\infty}b_n = 2 \wedge \left(\forall n \in \mathbb{N}_{+}\right)\left(b_n \leq a_n \leq c_n\right)\right)\Longrightarrow \lim_{n\to\infty}a_n = 2$$

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By differentiating $f(x)=\log ((2-1/x)^x)=x\log (2-1/x)=x(\log (2 x-1)-\log x)$ we see that $f'(x)>0$ for $x\geq 1. $ So the sequence $B_k=(2-1/k)^k=e^{f(k)}$ is increasing. Therefore $$2-(1/n)=B_n^{1/n}\leq ( \sum_{k=1}^nB_k)^{1/n}=a_n\leq (nB_n)^{1/n}=n^{1/n}B_n^{1/n}=n^{1/n}(2-(1/n)).$$

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