Show that the $\limsup_{n \to \infty}\sqrt[n]n$ $\leq 1$

Question

The question states: Show that the $\limsup_{n \to \infty}\sqrt[n]n$ $\leq 1$, and conclude that $\lim_{n \to \infty}\sqrt[n]n = 1$. (Hint: Assume that $\limsup_{n \to \infty}\sqrt[n]n$ $>1$ and arrive at a contradiction to the fact that $\sum_{n=1}^\infty n\theta^{n}<\infty$ for all $\theta \in(0,1)$.)

I am specifically trying to use the hint I was given to prove this statement as I know there are many alternate ways in showing it.

Assume that $\limsup_{n \to \infty}\sqrt[n]n$ $>1$,

I have let $b_n :=n\theta^{n}$ and tried using the Root Test to which I get: \begin{align} \limsup_{n \to \infty}b_n > \theta>0 \end{align} which doesn't help me. I don't see how any other test really helps me in this scenario as the hint makes me believe I need to use the $\limsup$ in some way. Any help is appreciated.

Answer 1

If $x = \liminf \sqrt[n]n > 1$, then $1 < \frac{x+1}2 < x$ and there are infinitly many $n$ such that $\sqrt[n]n\ge \frac{x+1}2$, that is, $n\left(\frac 2{x+1}\right)^n\le 1$. Now, sum them up, noting that $\theta = \frac 2{x+1} < 1$.

Answer 2

If you set $b_n = n\theta^n$ in the root theorem, you find $$\theta \limsup_{n\to\infty} \sqrt[n]{n} \leq 1$$ for all $0 < \theta < 1.$ (using the part of your hint not having to do with contradiction)

Divide both sides by $\theta > 0$: $$\limsup_{n\to\infty} \sqrt[n]{n} \leq 1/\theta$$

Now let $\theta \to 1^-$ to deduce the result.

Answer 3

Here is a different proof about the limit of $\sqrt[n]{n}$ using calculus. First, note that

$$f(x)=x^{\frac{1}{x}}$$

$$f'(x)=x^{\frac{1}{x}} \left(\frac{1}{x^2}-\frac{\log (x)}{x^2}\right).$$

Then the derivative is negative for $x>e$ as

$$f'(x)=x^{\frac{1}{x}} \left(\frac{1}{x^2}-\frac{\log (x)}{x^2}\right)<x^{\frac{1}{x}} \left(\frac{1}{x^2}-\frac{\log (e)}{x^2}\right)=0.$$

Thus, $f(x)$ is decreasing which implies the terms of $\sqrt[n]{n}$ are decreasing (at least for $n\geq 3$). In addition,

$$\sqrt[n]{n}\geq \sqrt[n]{1}=1.$$

Since we have a decreasing sequence that is bounded below by $1$, we may conclude that

$$\lim_{n\to\infty}\sqrt[n]{n}=L$$

exists and is greater than or equal to $1$. Now, suppose this limit is greater than $1$. Then there exists $\epsilon>0$ such that $\sqrt[n]{n}>1+\epsilon$ for all but a finite number of $n\in\mathbb{N}$. For $n\geq 2$, we may unravel this expression to get

$$\sqrt[n]{n}>1+\epsilon$$

$$n>(1+\epsilon)^n=\sum_{i=0}^n\binom{n}{i}\epsilon^i=1+n\epsilon+\frac{n(n-1)}{2}\epsilon^2+\cdots>\frac{n(n-1)}{2}\epsilon^2$$

$$2>(n-1)\epsilon^2$$

This is clearly false as we may choose $n\geq \frac{2+\epsilon^2}{\epsilon^2}$ to get

$$2>(n-1)\epsilon^2\geq \left(\frac{2+\epsilon^2}{\epsilon^2}-1\right) \epsilon^2=2.$$

Having reached a contradiction, we are assured that

$$\lim_{n\to\infty}\sqrt[n]{n}=1.$$