# Thread: A confusing sequence question

1. ## A confusing sequence question

Hello, I am completely lost with this question:

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

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

I see why $(1+ \frac{1}{n}) < n$ given that $n \geq 3$ , since $\lim (1+ \frac{1}{n}) = e < 3$
But that's pretty much what I got =(

2. Since you wrote " $\lim (1+ \frac{1}{n}) = e < 3$", I'm going to assume that you mean $(1+\frac{1}{n})^n$ instead.

We have the following bidirectional causal chain:

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

You should be able to do the rest now!