# Proof of the limit of abs [closed]

We have the expression to proof:

\begin{align*} \lim_{n \to \infty}|{\frac{1}{\sqrt[n^2]{n}}|=1} \end{align*}

What ways can we use for proving?

• What did you try? Why are you applying the absolute value to a sequence of numbers all of which are greater than $0$? Jun 22, 2019 at 5:34
• @JoséCarlosSantos oh, now I see the solution, sorry... We can replace
– Egor
Jun 22, 2019 at 5:38

Equality transformation: \begin{align*} {\frac{1}{\sqrt[n^2]{n}}}={\frac{1}{n^\frac{1}{n^{2}}}} \end{align*}

And replacing the variable:

\begin{align*} {\frac{1}{n}=u, u \to 0} \end{align*} Because \begin{align*} {n \to \infty} \end{align*}

Steps of replacing: \begin{align*} {\frac{1}{n^\frac{1}{n^{2}}}}={\frac{1}{n^{(\frac{1}{n})^2}}}={\frac{1}{n^{u^2}}}={(\frac{1}{n})}^{u^2}={u}^{u^2} \end{align*}

So we got easy expression:

\begin{align*} \lim_{u \to 0}|{{u}^{u^2}|=\lim_{u \to 0}|u^{u^2}|=\lim_{u \to 0}|u^{u}|=1} \end{align*}

• I believe $0^0$ is not defined. Jun 22, 2019 at 6:08
• @Epiksalad so, I editet this to limit of 0^0, what equals to 1
– Egor
Jun 22, 2019 at 7:12

$${\frac{1}{n^\frac{1}{n^{2}}}} = \exp\left(-\frac{\log n}{n^2}\right)$$ But $$\lim_{n \to \infty}\frac{\log n}{n^2} = 0$$ (either already known, or computed with L'Hospital)
so $$\lim_{n\to\infty}{\frac{1}{n^\frac{1}{n^{2}}}} = e^0 = 1$$

Let $$n >1$$;

$$a_n := n^{1/(n^2)}=(n^{1/n})^{1/n};$$

$$1\lt a_n \lt n^{1/n};$$

Hence $$\lim_{n \rightarrow \infty }a_n=1$$;

What can you say about $$\lim_{n \rightarrow \infty} b_n$$, where $$b_n=\dfrac{1}{a_n}$$.

Used: $$\lim_{n \rightarrow \infty} n^{1/n}=1$$.