# A confusing sequence question

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• Aug 17th 2010, 06:27 PM
akolman
A confusing sequence question
Hello, I am completely lost with this question:

Let $\displaystyle x_n = n^{\frac{1}{n}}$ for $\displaystyle n \in \mathbb{N}$.
1. Show that $\displaystyle x_{n+1} < x_n$ if and only if $\displaystyle (1+ \frac{1}{n}) < n$ and infer that the inequality is valid for $\displaystyle n \geq 3$. Conclude that $\displaystyle (x_n)$ is ultimately decreasing and that $\displaystyle x = \lim(x_n)$exists.

2. Use the fact that the subsequence $\displaystyle (x_{2n})$ also converges to $\displaystyle x$ to conclude that$\displaystyle x=1$.

I see why $\displaystyle (1+ \frac{1}{n}) < n$ given that $\displaystyle n \geq 3$ , since $\displaystyle \lim (1+ \frac{1}{n}) = e < 3$
But that's pretty much what I got =(
• Aug 17th 2010, 09:09 PM
Vlasev
Since you wrote "$\displaystyle \lim (1+ \frac{1}{n}) = e < 3$", I'm going to assume that you mean $\displaystyle (1+\frac{1}{n})^n$ instead.

We have the following bidirectional causal chain:

$\displaystyle \displaystyle \Leftrightarrow x_{n+1} > x_n$
$\displaystyle \displaystyle \Leftrightarrow n^{1/n} > (n+1)^{\frac{1}{n+1}}$
$\displaystyle \displaystyle \Leftrightarrow n^{\frac{n+1}{n}} > n+1$
$\displaystyle \displaystyle \Leftrightarrow n^{1+\frac{1}{n}} > n+1$
$\displaystyle \displaystyle \Leftrightarrow n.n^{\frac{1}{n}} > n+1$
$\displaystyle \displaystyle \Leftrightarrow n^{\frac{1}{n}} > \frac{n+1}{n} = 1+\frac{1}{n}$
$\displaystyle \displaystyle \Leftrightarrow n > \left(1+\frac{1}{n}\right)^n$

You should be able to do the rest now!