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According to the root test, we have that:

$\lim_{n\to\infty} \sqrt[n]{|a_n|} \left\{\begin{align}&>1 \implies \text{divergent}\\&=1\implies \text{???}\\&<1 \implies \text{convergence}\end{align}\right.$

But why doesn't it hold when you take the n-th power of both sides? Clearly $\lim_{n\to\infty} 1^n = 1$.

$\require{cancel} \color{green}{\large(}\lim_{n\to\infty} \sqrt[\cancel n]{|a_n|}\color{green}{\large)^\cancel{n}} = \lim_{n\to\infty} |a_n| = \left\{\begin{align}&>\color{green}(1\color{green}{)^n} = 1 \implies \text{divergent}\\&=\color{green}(1\color{green}{)^n} = 1 \implies \text{???}\\&<\color{green}(1\color{green}{)^n} = 1 \implies \text{convergence}\end{align}\right.$

Frank Vel
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1 Answers1

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Clearly $\lim_{n\to\infty} 1^n = 1$.

But $$\lim_{n\to\infty} a_n = 1$$ does not imply that $$\lim_{n\to\infty} (a_n)^n = 1$$ You can't just take the $n$-th power after taking the limit of the base; so in general: $$\lim_{n\to\infty} (a_n)^n \ne \left( \lim_{n\to\infty} a_n \right)^n$$

Take for example $a_n = 1+\tfrac{1}{n}$, then $a_n \to 1$ but $(a_n)^n \to e \ne 1^n = 1$.

In fact, this also explains why $1^\infty$ is one of the indeterminate forms.

StackTD
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  • $1^\infty$ is only indeterminate when $\lim_{n\to\infty} f(x)^g(x)$, $\lim_{n\to\infty} f(x) = 1$ and $\lim_{n\to\infty} g(x) = \infty$, but the right hand side it would just be $\lim_{n\to\infty} 1^n$ ($f(x) = 1, g(x) = n$), which would still be $1$? I can see it failing on the left hand side though. – Frank Vel Nov 06 '16 at 15:26