What are the values of the positive constant, $c$, for which $$ \sum_{n=2}^\infty \frac{1}{n(\ln n)^c}$$ is convergent or divergent?
I am a bit confused here, because usually the $c$ value is given, so I am not really sure how to approach this.
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$\begingroup$Using the integral test, for a set $a$, we see $$\lim_{b\to\infty}\int_{a}^{b}\frac{1}{x\log(x)^{c}}dx=\lim_{b\to\infty}\left(\frac{\log(b)^{1-c}}{1-c}-\frac{\log(a)^{1-c}}{1-c}\right)$$ which goes to infinity if $c\leq 1$ and converges if $c>1$. Thus, $$\sum_{n=2}^{\infty}\frac{1}{n\log(n)^c}$$ converges if and only if $c>1$. I discarded $n=1$ since $1*\log(1)^c=0$ which introduces dividing by $0$.
$\endgroup$ 2 $\begingroup$First, comparison to $\sum \frac{1}{n \ln n}$ tells you that the series diverges if $c \le 1$. With $c > 1$, check that you can apply Cauchy's condensation test. The test gives you
$$ \frac{2^n}{2^n (\ln 2^n)^c} = \frac{1}{n^c (\ln 2)^c}, $$
which compares nicely with a $p$-series.
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