Question regarding a series that contains logarithms $$\sum_{n=1}^\infty\bigl(3\log(n^2+1)-2\log(n^3+1)\bigr)$$
I tried the limit comparison test with $\sum_{n=1}^\infty\frac1{n^2}$ and then applied L'Hopital's Rule. What I ended up was a limit which was a real number and as the series of $\frac1{n^2}$ converges, my original series converges as well. However, I was wondering if there is another way of solving the problem, one that involves fewer computations.
 A: $$3\ln(n^2+1)-2\ln(n^3+1)=\ln\left( \frac{(n^2+1)^3}{(n^3+1)^2}\right)=\ln\left( \frac{n^6+3n^4+3n^2+1}{n^6+2n^3+1}\right)$$
$$=\ln\left( 1+\frac{3n^4-2n^3+3n^2}{n^6+2n^3+1}\right) \sim \frac{3n^4-2n^3+3n^2}{n^6+2n^3+1} \sim \frac{3}{n^2}$$
Therefore the series converges.
A: For each $n\in\Bbb N$, you have\begin{align}3\log(n^2+1)-2\log(n^3+1)&=\log\left(\frac{(n^2+1)^3}{(n^3+1)^2}\right)\\&=\log\left(\frac{\left(1+\frac1{n^2}\right)^3}{\left(1+\frac1{n^3}\right)^2}\right).\end{align}But, near $0$, you have$$\log\left(\frac{(1+x^2)^3}{(1+x^3)^2}\right)=3x^2-2x^3+\cdots$$and therefore$$\lim_{n\to\infty}\frac{3\log(n^2+1)-2\log(n^3+1)}{\frac1{n^2}}=3.$$Can you take it from here?
A: The series at hand can be rewritten as
$$\sum_{n=1}^\infty\left[ 3(\log(n^2+1) - 2\log(n)) - 2(\log(n^3+1)-3\log(n))\right]$$
For any $\alpha > 1$, if we apply MVT to $\log(x)$ over interval $(n^\alpha,n^\alpha+1)$, we find there is a $\xi \in (0,1)$ such that
$$\log(n^\alpha+1) - \alpha\log(n)
= \log(n^\alpha+1) - \log(n^\alpha) = \frac{1}{n^\alpha + \xi} \le \frac{1}{n^\alpha}$$
Since these terms are non-negative and $\displaystyle\;\sum_{n = 1}^\infty \frac{1}{n^\alpha} = \zeta(\alpha) < \infty$, series of the form
$$\mathcal{S}_\alpha \stackrel{def}{=} \sum_{n=1}^\infty (\log(n^\alpha+1) - \alpha\log(n))$$
converges. The series at hand is a linear combination of $\mathcal{S}_2$ and $\mathcal{S}_3$ and hence converges.

As a side note, I originally misunderstood the question to one about the evaluating the series. I did evaluate it before realizing the error. I will
leave the derivation below in case anyone cares.
When $\alpha$ is a positive integer $\ge 2$, we can express
$\mathcal{S}_\alpha$ in terms of gamma function. We will only evaluate $\mathcal{S}_2$ and $\mathcal{S}_3$ for demonstration.
Recall for and $z \in \mathbb{C}$, we have following infinite product expansion: $$\frac{1}{\Gamma(z+1)} = e^{\gamma z}\prod_{n=1}^\infty \left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$$
where $\gamma$ is the Euler–Mascheroni constant. Furthermore, gamma function satisfies following functional relation:
$$\Gamma(1+z) = z\Gamma(z)\quad\text{ and }\quad \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$
For $\mathcal{S}_2$, we have
$$\begin{align}
\mathcal{S}_2 &= 
\sum_{n=1}^\infty(\log(n^2 + 1) - 2\log(n))
= \sum_{n=1}^\infty\log\left(1 + \frac{1}{n^2}\right)\\
&= \log\left[\prod_{n=1}^\infty\left(1 + \frac{1}{n^2}\right)\right] = \log\left[\prod_{n=1}^\infty(1 + \frac{i}{n})(1-\frac{i}{n})\right]\\
&\stackrel{(*)}{=} \log\left[
\left(e^{\gamma i}\prod_{n=1}^\infty(1 + \frac{i}{n})e^{-\frac{i}{n}}\right)
\left(e^{-\gamma i}\prod_{n=1}^\infty(1 - \frac{i}{n})e^{\frac{i}{n}}\right)
\right]\\
&= -\log[\Gamma(1+i)\Gamma(1-i)]\\
&= -\log[ i\Gamma(i)\Gamma(1-i)]\\
&= -\log\left[\frac{\pi i}{\sin(\pi i)}\right]\\
&= \log\sinh(\pi) - \log\pi
\end{align}
$$
In about derivation, step $(*)$ is possible because $i + (-i) = 0$. This allows us to push extra factors (which cancel among each other) into the product and make it look like that in gamma function's infinite product expansion.
For $\mathcal{S}_3$, the situtation is similar. Let $\omega = e^{\frac{2\pi i}{3}}$ be the cubic root of unity. We will use the fact $\omega^2 + \omega + 1$ to preform the same trick.
$$\begin{align}\mathcal{S_3} 
&= \sum_{n=1}^\infty \log(n^3+1) - 3\log(n) = \sum_{n=1}\log\left(1 + \frac{1}{n^3}\right)\\
&= \log\left[\prod_{n=1}^\infty\left(1 + \frac{1}{n^3}\right)\right]
= \log\left[\prod_{k=0}^2\prod_{n=1}^\infty\left( 1 + \frac{\omega^k}{n}\right)\right]\\
&= \log\left\{\prod_{k=0}^2 \left[e^{\gamma \omega^k}\prod_{n=1}^\infty\left(1 + \frac{\omega^k}{n}\right) e^{-\frac{\omega^k}{n}}\right]\right\}\\
&= -\log(\Gamma(1+1)\Gamma(1+\omega)\Gamma(1+\omega^2))\\
&= -\log(\Gamma(1+\omega)\Gamma(-\omega))\\
&= \log \sin(-\pi\omega) -\log\pi\\
&= \log \cosh\left(\pi\frac{\sqrt{3}}{2}\right) - \log\pi
\end{align}
$$
Combine these, we find the series at hand converges to
$$\begin{align}3\mathcal{S}_2 - 2\mathcal{S}_3 &= 3\log\sinh(\pi) - 2\log\cosh(\frac{\pi\sqrt{3}}{2}) - \log\pi\\ &\approx 2.131247119366045
\end{align}$$
A: Similar to TheSilverDoe's answer, we have that eventually
$$3\log(n^2+1)-2\log(n^3+1)=\log\left(\frac{(n^2+1)^3}{(n^3+1)^2}\right)=\ln\left( 1+\frac{3n^4-2n^3+3n^2}{n^6+2n^3+1}\right) \le$$
$$\le \frac{3n^4-2n^3+3n^2}{n^6+2n^3+1} \le \frac{4n^4}{\frac12n^6}=8\frac1{n^2}$$
and we can conclude by direct comparson test.
