We have to prove that $n^{\frac{1}{n}}$ converges to $1$. I have proved it using the binomial theorem where we can substitute $(1+t)$ in place of $n$ and proceed forward. However along with the question another approach was mentioned where we can make use of the Monotone Convergence Theorem. For this approach, I proved that $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing and using $a_n<a_{n+1}$ I proved that $n^{\frac{1}{n}}$ is decreasing after the third term. Also it is bounded as all terms are greater than $0$. So by MCT it should be convergent. But I am not able to evaluate the limit and am only able to prove its existence here. Please help.
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Please take a few minutes to format your question: see https://math.stackexchange.com/help/notation. – Théophile Dec 07 '20 at 04:35
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1Why the downvote? – Alann Rosas Dec 07 '20 at 04:40
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I think you mean $(1+1/n)^n$, not $(n+1/n)^n$. – marty cohen Dec 07 '20 at 05:03
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@A.E.Rosas . In 5 years I have never seen anyone except myself give a reason for a downvote. I rarely do, but I always explain it. – DanielWainfleet Dec 07 '20 at 05:07
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@martycohen yes you are correct. Sorry it was a typo – V2002 Dec 07 '20 at 05:13
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After you have shown that $L=\lim_{n\to \infty}n^{1/n}$ exists, and is positive (because every $n^{1/n}\ge 1$), then $$0<L=\lim_{n\to\infty}(2n)^{1/2n}=\lim_{n\to\infty}2^{1/2n}(n^{1/n})^{1/2}=L^{1/2}$$ because $2^{1/2n}\to 1.$
DanielWainfleet
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Easiest proof:
By Bernoulli,
$(1+n^{-1/2})^n \ge 1+n^{1/2} \gt n^{1/2} $.
Raising to $2/n$ power,
$n^{1/n} \lt (1+n^{-1/2})^2 =1+2n^{-1/2}+n^{-1} \lt 1+3n^{-1/2} \to 1 $.
marty cohen
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$n^{1/n}=e^{\ln n^{1/n}}=e^{\ln n/n}$. Now exp is continuous, and $\ln n/n\to0$. So the limit is $e^0=1$.
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$$a_n = n^{1/n} \implies \ln a_n = \frac{\ln n}{n} \implies \lim\ln a_n = 0 \implies \ln \lim a_n = 0 \implies \lim a_n = e^0 = 1$$
VIVID
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