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\noindent {\bf Question}

\noindent Determine whether the infinite series
\[ \sum_{n=3}^\infty {1\over n \ln(n)} \]
converges or diverges.  (You do not need to evaluate the sum of
the series in the case that it converges.)

\medskip

\noindent {\bf Answer}

\noindent Since the terms in the series are all positive, we may
use the integral test, with $f(x) =\frac{1}{x\ln(x)}$.  This
function is continuous for $x \ge 3$ and is decreasing, since
$f'(x) = -\frac{\ln(x) +1}{x^2\ln^x(x)} <0$ for $x \ge 3$.  Then,
the series converges if and only if the improper integral
$\int_3^\infty \frac{1}{x\ln(x)} {\rm d}x
=\lim_{M\rightarrow\infty} \int_3^M \frac{1}{x\ln(x)} {\rm d}x$
converges. Calculating, we see that
\[ \lim_{M\rightarrow\infty} \int_3^M \frac{1}{x\ln(x)} {\rm d}x
=\lim_{M\rightarrow\infty} \ln(\ln(x))\left|_3^M \right.
=\lim_{M\rightarrow\infty} (\ln(\ln(M)) -\ln(\ln(3)) =\infty. \]
Since the integral diverges, the series diverges.


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