Finding a constant where an improper integral converges I am trying to find the interval for the value $p$ where the following integral converges (log is the natural logarithm):
$$\int_1^{\infty} \frac{\log{x}}{x^p} \,dx$$ 
So far I've used integration by parts to reduce the integral to the limit:
$\displaystyle \lim_{a\to\infty} \frac{\log{x} - p \log{x} - 1}{x^{p-1}(1-p)^2}$. 
My intuition tells me that $p > 2$, but I am struggling to find a good argument. 
A push in the right direction would be appreciated!
 A: Recall that the logarithm function satisfies the inequality 
$$\log(x)\le x-1<x \tag 1$$
for $x>0$.
Then, since $\log(x^{\alpha})=\alpha \log(x)$, we have from $(1)$ for $\alpha>0$
$$\begin{align}
\alpha \log(x)&< x^{\alpha}\\\\
\log(x)&<\frac{x^{\alpha}}{\alpha} \\\\
\frac{\log(x)}{x^p}&< \frac{x^{\alpha-p}}{\alpha}
\end{align}$$
Therefore, for any $p>1$ we can take $\alpha>0$ small enough so that $p-\alpha>1$.  Therefore the integral of interest converges for all $p>1$ by the comparison test.

Since $\log(x)>1$ for $x>e$, then $\frac{\log(x)}{x^p}>\frac{1}{x^p}$ for $x>e$ and the integral diverges for $p\le 1$.

We can evaluate the improper integral applying integration by parts with $u=\log(x)$ and $v=x^{1-p}/(1-p)$.  Proceeding accordingly, we obtain
$$\begin{align}
\lim_{L\to \infty}\int_1^L \frac{\log(x)}{x^p}\,dx&=\lim_{L\to \infty}\left(\left. \frac{x^{1-p}\log(x)}{1-p}\right|_{1}^{L}\right)+\frac{1}{p-1}\lim_{L\to \infty }\int_1^L\frac{1}{x^p}\,dx\\\\
&=\frac{1}{(p-1)^2}
\end{align}$$
A: It's clear that $p$ must be positive. Use integral test as the integrand is monotone decreasing,
$$
\sum_{n=1}^\infty{\ln n\over n^p}
$$
The above series converges iff by Cauchy condensation test the series below converges
$$
\sum_{n=1}^\infty{n2^n\over2^{np}}
$$
which converges if $p>1$ by root test.
