Finding what $p$ does the series: $\sum_{n=1}^{\infty}\frac{\ln n}{n^p}$ converges 
For what $p$ does the series: $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^p}$ converges?

My attempt:
I wanted to use the limit comparison test and compare it with $\frac 1 {n^p}$ but it doesn't work since $\ln n \to \infty$.
So I tried applying the condensation test twice: 
$\large\sum \frac {n 2^n \ln2}{2^{np}}=\sum \frac { 2^{2^n+2n} \ln2}{2^{2^np}}$
Now we want the denominator to be larger than the numerator so we want to solve: $2^n+2n<2^np\to 1+2n/2^n < p$
Now since $2n/2^n\to0$ then for large enough $n$ we'll have $1<p$. 
Does this approach alright? Is there another way?
 A: Hint. You may observe that your series behaves like the integral$$
\int_1^{\infty} \frac{\ln x}{x^p}dx
$$ then, integrating by parts, we get
$$
\int_1^{\infty} \frac{\ln x}{x^p}dx=-\frac{1}{(p-1)x^{p-1}}\ln x+\frac{1}{(p-1)}\int_1^{\infty} \frac{1}{x^p}dx, \quad p \neq1,
$$the last integral being convergent if and only if $p>1$ (the case $p=1$ is directly seen to be divergent).
A: Since $\dfrac{\ln n}{n^p} \ge \dfrac{1}{n^p}$ for all $n \ge 3$ the direct comparison test tells you that the series is divergent for all $p \le 1$.
If $p > 1$, use the limit comparison test with the series $\displaystyle \sum_{n=1}^\infty \frac{1}{n^{(p+1)/2}}$. This series converges, and since $$\lim_{n \to \infty} \frac{\ln n}{n^p} \cdot n^{(p+1)/2} = \lim_{n \to \infty} \frac{\ln n}{n^{(p-1)/2}} = 0$$
so does the original series.
A: Why twice? That only makes things messier, imo:
$$\frac n{2^{n(p-1)}}\log 2$$
and now apply the $\;n$-th root test:
$$\frac{\sqrt[n]n}{2^{p-1}}\sqrt[n]{\log2}\xrightarrow[n\to\infty]{}\frac1{2^{p-1}}$$
A: You can use the integral test since $f(x) = \frac{\ln x}{x^p}$ is decreasing for $x > e^{1/p}$. Then via the substitution $t = \ln x, dt = x^{-1}dx, dx = e^tdt$
$$
\int_1^\infty \frac{\ln x}{x^p} dx = \int_1^\infty t \cdot e^{(1-p)t} dt = (1-p)^{-2}e^{t(1-p))}(t(1-p)-1)|_{t=0}^{t=\infty}
$$
and this is finite if and only if $\boxed{p > 1}$.
