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\noindent {\bf Question}

\noindent Determine whether the sequence
\[ \left\{ a_n = \frac{\left( \frac{2}{3} \right)^n}{2 - n^{1/n}}
\right\} \] converges or diverges.  If the sequence converges,
determine its limit.

\medskip

\noindent {\bf Answer}

\noindent We know that $\lim_{n\rightarrow\infty} (\frac{2}{3})^n
=0$, since $\frac{2}{3} <1$.  Hence, we need to evaluate
$\lim_{n\rightarrow\infty} n^{1/n}$:  start by writing
\[ n^{1/n} = \exp(\ln(n))^{1/n} = \exp\left( \frac{\ln(n)}{n}
\right). \] Since $\lim_{n\rightarrow\infty} n^{1/n} =\exp\left(
\lim_{n\rightarrow\infty} \frac{\ln(n)}{n} \right)$, and since
$\lim_{n\rightarrow\infty} \frac{\ln(n)}{n}$ has the indeterminate
form $\frac{\infty}{\infty}$, we may use l'Hopital's rule to
evaluate:
\[ \lim_{n\rightarrow\infty} \frac{\ln(n)}{n}
=\lim_{n\rightarrow\infty} \frac{ \frac{1}{n}}{1} =0, \] and so
\[ \lim_{n\rightarrow\infty} n^{1/n} = \exp\left(
\lim_{n\rightarrow\infty} \frac{\ln(n)}{n} \right) = e^0 =1.\]
Hence, the original limit can be evaluated using the arithmetic of
limits:
\[ \lim_{n\rightarrow\infty} \frac{\left( \frac{2}{3} \right)^n}{2 -
n^{1/n}} = \frac{0}{2-1} =0, \] and so the sequence converges to
$0$.


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