Answer
The sequence converges,
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=0$
Work Step by Step
Th.9.1:
"Let $L$ be a real number.
Let $f$ be a function of a real variable such that $\displaystyle \lim_{x\rightarrow\infty}f(x)=L$.
If $\{a_{n}\}$ is a sequence such that $f(n)=a_{n}$ for every positive integer $n$,
then $\displaystyle \lim_{n\rightarrow\infty}a_{n}=L$. "
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We define $f(x)=\displaystyle \frac{\ln(x^{3})}{2x}$, for which$ f(n)=a_{n}$ and
$\displaystyle \lim_{x\rightarrow\infty}f(x)$ is of the form $\displaystyle \frac{\infty}{\infty}$,
found by L'Hopital's rule,
$[\ln(x^{3})]^{\prime}=$ ...chain rule...$ =\displaystyle \frac{1}{x^{3}}\cdot 3x^{2}=\frac{3}{x}$
$[2x]^{\prime}=2$
$\displaystyle \lim_{x\rightarrow\infty}f(x)=\frac{3}{2x}=$
--- when $ x\rightarrow\infty$, the numerator is constant and the denominator is very large, so
$\displaystyle \lim_{x\rightarrow\infty}f(x)=0$
Thus,
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=\lim_{x\rightarrow\infty}f(x)=0$.
