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.


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 $$\begin{align}3\mathcal{S}_2 - 2\mathcal{S}_3 &= 3\log\sinh(\pi) - 2\log\cosh(\frac{\pi\sqrt{3}}{2}) - \log\pi\\ &\approx 2.131247119366045 \end{align}$$

  • $\begingroup$ I just started Calculus so the part with the Gamma function is unknown territory for me, but I'll save it in case I'll need to evaluate it someday. Also, I love your demonstration with the MVT! I would have never thought about decomposing the terms of the series like that and then using it. Do you use it often in evaluating the convergence of different series? $\endgroup$ – abigmistake Oct 28 '20 at 15:12
  • $\begingroup$ @abigmistake about decomposing the terms, the answer is "sort of" a yes. When you want to analysis something complicated, one strategy is decomposing the "something" into smaller structural unit (note not individual terms) and attack each unit individually. By breaking into smaller unit, you reduce the influences among them. This make it easier in analysing the target. This tactics isn't limited to math, it also work in real world problems. $\endgroup$ – achille hui Oct 28 '20 at 18:28

$$3\ln(n^2+1)-2\ln(n^3+1)=\ln\left( \frac{(n^2+1)^3}{(n^3+1)^2}\right)=\ln\left( \frac{n^6+3n^4+3n^2+1}{n^6+2n^3+1}\right)$$ $$=\ln\left( 1+\frac{3n^4-2n^3+3n^2}{n^6+2n^3+1}\right) \sim \frac{3n^4-2n^3+3n^2}{n^6+2n^3+1} \sim \frac{3}{n^2}$$

Therefore the series converges.


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?

  • $\begingroup$ Thank you very much! $\endgroup$ – abigmistake Oct 25 '20 at 10:52

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.