Convergence of $\sum_{n=1}^{\infty}\frac{\log(n)}{n^a+b}$. Is it true that 
$$\sum_{n=1}^{\infty}\frac{\log(n)}{n^a+b} $$
is convergent for $a >1$ and $b \geq 0$? If not, for what (if any) values of $a,b$ does it converge?
Trying to apply the well-known inequality $\log(x) \leq x$ directly doesn't seem to be useful. It may also not be clear how the inequality $\log(x) \leq x^{\alpha}$ for $\alpha>0$, which holds eventually, may be used here. For instance, taking $\log(x) \leq x^a$ yields $\log(x)/x^a \leq 1$ eventually, which is no good. 
Since we are assuming $a>1$, trying $\log(x) \leq x^{a-1}$ could be a reasonable attempt, but this leaves us at $\log(x)/x^a \leq 1/x$ holding eventually, which will amount to comparing with the harmonic series.
 A: It is indeed true that
$$\sum_{n=1}^{\infty}\frac{\log(n)}{n^a+b}$$
converges for every $a>1$ and $b \geq 0$. $^{(1)}$ (Note that it obviously diverges if $a=1$ by comparison or limit comparison in the case $b \neq 0$ with the harmonic series.)
One way to see this is by using the fact that for every $\alpha>0$, it is true that $\log(x)<x^\alpha$ eventually.$^{(2)}$
Given that, pick $a'$ such that $a>a'>1$. Therefore $a-a'>0$ and $a'>1$. It follows that $\log(n)<n^{a-a'}$ eventually, which implies that
$$\frac{\log(n)}{n^a+b} \leq \frac{\log(n)}{n^a} \leq \frac{1}{n^{a'}} $$
eventually. Comparing with the $p$-series with $p=a'$ thus yields convergence. (It may be worth noticing that this small leeway of taking $a'$ slightly greater than $1$ is essentially what fixes the issue of ending up with a comparison with the harmonic series as in the OP.)
Another way is by employing the limit comparison test. This ends up being essentially the same as above. Pick $p$ such that $a>p>1$. Thus, by considering the $p$-series as the source of comparison, we get that the series is convergent if the limit
$$\frac{ \log(n)/(n^a+b)}{1/n^p}=\frac{\log(n)}{n^{a-p}+\frac{b}{n^p}}$$
is finite. The right side is smaller than $\frac{\log(n)}{n^{a-p}}=n^{p-a}\log(n)$, which converges to $0$ as $n \to +\infty$, thus yielding the result.
Yet another way is by using the Cauchy condensation test. First we establish that $\frac{\log(x)}{x^a}$ is eventually decreasing. Its derivative is given by
\begin{align*}
\left(\frac{\log(x)}{x^a}\right)'&=\frac{x^{a-1}-\log(x)ax^{a-1}}{x^{2a}} \\
&= \frac{x^{a-1}(1-a\log(x))}{x^{2a}}.
\end{align*}
Since $1-a\log(x)$ is eventually negative and all other factors are positive, it follows that the derivative is also eventually negative. We can thus apply the Cauchy condensation test. By doing so, we know that the original series is convergent if and only if
$$\sum_{n=1}^{\infty}2^n\cdot\frac{\log(2^n)}{2^{na}+b}=\sum_{n=1}^{\infty}\log(2)\cdot\frac{2^nn}{2^{na}+b}$$
is also convergent. By comparison and taking away the constant $\log(2)$, it suffices to analyse convergence of
$$\sum_{n=1}^{\infty}\frac{2^nn}{2^{na}}=\sum_{n=1}^{\infty}\frac{n}{2^{n(a-1)}}.$$
By the ratio test, this series converges if $a>1$ and thus we conclude what we wanted.

$^{(1)}$ Requiring $b \geq 0$ here is not essential. This is just to avoid possibly having to disregard some undefined term of the sum. The result is true if we do that.
$^{(2)}$ This is a well-known inequality, but for the sake of completeness, a proof can be given as follows: we know that $\frac{e^{\alpha x}}{x} \to +\infty$, since $e^{\alpha x} \geq \frac{\alpha^2x^2}{2}$. It follows that $e^{\alpha x} \geq x$ eventually, and thus $e^{\alpha x^{\alpha}} \geq x^{\alpha}$ eventually. Taking the logarithms yields the desired inequality. One could also apply L'Hôpital's rule to $\frac{\log(x)}{x^{\alpha}}$ if one wishes to.
A: Note that $\log(x)<x-1$ for all $x>0$.  In addition, $\log(x^\alpha)=\alpha \log(x)$.  
Hence, we deduce that for $\alpha>0$
$$\log(x)<\frac{x^\alpha -1}{\alpha}<\frac{x^\alpha}{\alpha}$$
Then, we can write for $\alpha>0$ and $b>0$
$$\frac{\log(n)}{n^a+b}<\frac{n^\alpha}{\alpha(n^a+b)}<\frac1\alpha n^{\alpha-a}$$
Choosing any $0<\alpha<a-1$, we see that 
$$\sum_{n=1}^\infty \frac{\log(n)}{n^a+b}$$
converges whenever $a>1$ and $b>0$.
