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How to determine convergence and divergence of $a_n =\sqrt[\Large^n]c$ ?

omidh
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    You can understand the behavior of this sequence very precisely by writing $\sqrt[n]{c} = \exp \left( \frac{\log c}{n} \right)$. This tells you not only what the limit is but also the leading order of the error, and in fact an asymptotic expansion to all orders in $\frac{1}{n}$. – Qiaochu Yuan Dec 14 '20 at 06:26

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Assume $c > 0$. Further assume $c > 1$ because if $c < 1$ we can put $c = \dfrac{1}{b} < 1, b > 1$. Thus put $c = a+1, a > 0$, then $ 1 < a_n = (1+a)^{\frac{1}{n}}= \sqrt[n]{1\cdot 1\cdot 1\cdots (1+a)} < \dfrac{1+1+\cdots + 1+(1+a)}{n}= \dfrac{n+a}{n}= 1+\dfrac{a}{n}$. Thus $a_n \to 1$ by sandwich theorem. If $c = 1$, the limit is $1$.