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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?

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

3 Answers 3

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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*}

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  • $\begingroup$ I believe $0^0$ is not defined. $\endgroup$ Jun 22, 2019 at 6:08
  • $\begingroup$ @Epiksalad so, I editet this to limit of 0^0, what equals to 1 $\endgroup$
    – Egor
    Jun 22, 2019 at 7:12
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$$ {\frac{1}{n^\frac{1}{n^{2}}}} = \exp\left(-\frac{\log n}{n^2}\right) $$ But $$ \lim_{n \to \infty}\frac{\log n}{n^2} = 0 $$ (either already known, or computed with L'Hospital)
so $$ \lim_{n\to\infty}{\frac{1}{n^\frac{1}{n^{2}}}} = e^0 = 1 $$

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

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