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given a sequence $a_n=a^{\frac{1}{n}}$ for $n\in\mathbb{N}^*$, $a\in\mathbb{R},a>1$ then proof that $\lim\limits_{n\to+\infty}a_n=1$ by definition.

proof:

given $a_n=a^{\frac{1}{n}}$ for $a\in\mathbb{R},a>1$.

for $n\in\mathbb{N}^*,n+1>n\Rightarrow \frac{1}{n+1}<\frac{1}{n}$ and because $a>1$ we gets $a^{\frac{1}{n+1}}<a^{\frac{1}{n}}$, since $\frac{1}{n+1}>0$, then $1=a^{0}<a^{\frac{1}{n+1}}<a^{\frac{1}{n}}\Rightarrow 1<a_{n+1}<a_{n}$.

then we proof that $\lim\limits_{n\to+\infty}a_n=1$ we need to show that $\forall\epsilon>0,\exists N,\forall n>N,|a_n-1|<\epsilon$

then for $\epsilon>0$, choose $N=\frac{1}{\log_a(\epsilon+1)}$, since $a>1$ we gets that $$\forall n>N,1<a^{\frac{1}{n}}<a^\frac{1}{N}\Rightarrow0<a^{\frac{1}{n}}-1<a^{\frac{1}{N}}-1\Rightarrow |a^{\frac{1}{n}}-1|<|a^{\frac{1}{N}}-1|=|a^{\log_a(\epsilon+1)}-1|=|\epsilon+1-1|=\epsilon\Rightarrow |a^{\frac{1}{n}}-1|<\epsilon$$

wich implies that $\lim\limits_{n\to+\infty}a_n=1\square$

my proof to the limit is correct?

Martin Sleziak
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cand
  • 2,266

2 Answers2

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Yes. your proof is correct. for proof by epsilon and delta, you must find a delta for given epsilon; that you find a $N$ for given epsilon.

Farhad
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You didn't need the first paragraph in your proof, just the second paragraph. In the second paragraph, you cannot write $N = \frac{1}{\log_a(\epsilon + 1)}$ because $\frac{1}{\log_a(\epsilon + 1)}$ is not necessarily an integer. However, you can let $N > \frac{1}{\log_a(\epsilon + 1)}$. The rest of the proof is fine.

kobe
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