proof that the series converges? I just need to make sure that I do it correctly 



Thanks in advance,
Hint

$$
\frac{|\log x|}{1+x^2}\leq |\log x|
$$
You can use the fact $\ln(x) &lt; x$ which makes it easier 

$$ \int_{0}^{1}\frac{\ln(x)}{1+x^2}dx &lt;  \int_{0}^{1}\frac{x}{1+x^2}dx=\frac{1}{2}\ln(1+x^2)|_{x=0}^{1}=\dots. $$

Note: Just note that, $ \frac{\ln(x)}{1+x^2} $ is Riemann integrable (in the extended sense if you wish and $x\sim 0$ $\frac{\ln(x)}{1+x^2}\sim \ln(x)$ ) and by the property of the Riemann integral, for $f$ and $g$ Riemann integrable functions such that 

$$ f \leq g\quad  \forall x\in[a,b] \implies \int_{a}^{b} f(x)dx \leq \int_{a}^{b} g(x)dx, $$

then above answer is true 

Another approach: Note that the integrand has a singularity at $x=0$, so we have

$$ x\sim 0 \implies \frac{\ln(x)}{1+x^2} \sim \ln(x) $$

and using the fact 

$$ \int_{0}^{1}\ln(x)dx &lt; \infty $$

the original integral converges.