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To show that sequence $\left(1+{1 \over 3n}\right)^n$ converges using cauchy criteria.

For a given $\epsilon$, $\exists$ natural number $k$ such that, $$|a_m-a_n|<\epsilon, \forall m,n \ge k$$.

for given sequence, $|a_m-a_n|$=$\left|(1+{1 \over 3m})^m-(1+{1 \over 3n})^n \right|$=$\left|1+3+{m(m-1)\over 2}{1 \over{(3m)^2}}. . .|-|1+3+{n(n-1)\over 2}{1 \over{(3n)^2}}. . .\right|$

what after this?

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By AM-GM inequality, for all $n\in\mathbb N$, \begin{align*} \textstyle \left(\frac{1}{1+1/(3n)}\right)^{(3n+1)/(3n+2)}=\left(\frac{3n}{3n+1}\right)^{(3n+1)/(3n+2)}\leq\frac{1+(3n+1)(3n/(3n+1))}{3n+2}=\frac{3n+1}{3n+2}=\frac{1}{1+1/(3n+1)}. \end{align*} Thus, $a_n:=\left(\frac{1}{1+1/(3n)}\right)^{3n+1}$ is a decreasing sequence. Since $a_n$ has a lower bound $0$, it is a Cauchy sequence. Then your sequence $a_n^{1/(3+1/n)}$ is also a Cauchy sequence.

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