Comprehensive Analysis of the Evaluation of an Infinite Logarithmic Summation

Question

Comprehensive Analysis of the Evaluation of an Infinite Logarithmic Summation

Epistemic Framework As I pondered the quintessential nature of the intricate mathematical fabric of the universe, contemplating its underlying structures and elegant symmetries, I found myself captivated by the challenge of uncovering a closed-form expression for the following summation.

$$\sum_{k=1}^{\infty}\frac{\log(k)}{k^2+k} = \sum_{k=1}^{\infty}\frac{\log(k)}{k(k+1)} $$

I feel compelled to emphasize that $\log(x)$ denotes the logarithm to the base of the mathematical constant $e$. Motivated by my insatiable curiosity and a desire to unravel the mysterious form of this infinite series, I embarked on an analytical journey to the fundamental nature of this chimera. I shall include my exhaustive attempts below.

My Sincere Mathematical Exertions I first observed that the denominator $k^2 + k$ can be factored, namely $k^2 + k = k(k+1)$. Therupon, the general term of the series becomes $\log(k)/k(k+1)$. I imbued the latter mathematical expression with my terse experience concerning partial fraction disintegration; I deduced that the general term assumes an extraordinarily novel form, to which I shall notate below these lines of text. $$\frac{\log(k)}{k(k+1)} = \log(k)\Big(\frac1k - \frac{1}{k+1} \Big) = \frac{\log(k)}{k} - \frac{\log(k)}{k+1}$$

Substituting the decomposed form back into this infinite series, we obtain the following \begin{align} &\sum_{k=1}^{\infty} \Big(\frac{\log(k)}{k} - \frac{\log(k)}{k+1}\Big) = \Big(\frac{\log(1)}{1} - \frac{\log(1)}{2}\Big) + \Big(\frac{\log(2)}{2} - \frac{\log(2)}{3}\Big) + \cdots \\ &= \log(1) - \frac{\log(1)}{2} + \frac{\log(2)}{2} - \frac{\log(2)}{3} + \frac{\log(3)}{3} - \frac{\log(3)}{4} + \cdots \\ &= \log(1) + \sum_{k=2}^{\infty} \Big(\frac{\log(k)}{k} - \frac{\log(k-1)}{k}\Big) \end{align}

I reanimated and elucidated this form, enabled by the magnificent theorem that $\log(1) = 0$. Ergo and for the importance of brevity, I found my eyes captivated on a beautiful and innovative structure.

$$\sum_{k=2}^{\infty} \Big(\frac{\log(k)}{k} - \frac{\log(k-1)}{k}\Big)$$

Unfortunately, after exerting my mental capabilities, I was unable to obtain a convincing path forward. An ancient proverb that comes to mind is "All our dreams can come true, if we have the courage to pursue them" (Walt Disney). Alas! With the use of a futuristic tool called "Wolf and alpha," I was able to probe the this mathematical intrigue further. The partial sums of the partial parts may concern themselves with $\gamma_n$, the $n$-th Stieltjes constant. I shall provide an internet-based linkage of a source that pertains to the Stieltjes constants here: the aforementioned linkage.

Culminating deliberations I do sincerely hope that my attempts appear sincere; as of this moment, this problem is the greatest trouble to my mind. Any insight that may pertain to the monumental discovery of a closed form for this series would be tremendous. I do hope that this online-internet based posting was concise and clear.

Sincerely yours, CommunityBot.

Answer 1

I'm just an undergraduate, I'm not sure if what I'm thinking is wrong. But there seems to be some very basic problems here? I want to make a comment, but I don't have permission, I can only write it here. Here are some of my thoughts on your original derivation. If you can sort out these very basic errors and knowledge, and then modify your question slightly, maybe you can get any real meaningful help on this issue.

When splitting the original series into $\sum_{k=1}^{\infty} \left( \frac{\ln(k)}{k} - \frac{\ln(k)}{k+1} \right)$, you’re treating it as $\sum_{k=1}^{\infty} \frac{\ln(k)}{k} - \sum_{k=1}^{\infty} \frac{\ln(k)}{k+1}$. Problem is, both $\sum_{k=1}^{\infty} \frac{\ln(k)}{k}$ and $\sum_{k=1}^{\infty} \frac{\ln(k)}{k+1}$ diverge—think integral test, $\int_1^\infty \frac{\ln x}{x} dx$ blows up. Subtracting two divergent series is undefined unless the whole thing converges and the split holds in the limit. It doesn’t work like that here.

Then you rewrite it as $\sum_{k=2}^{\infty} \left( \frac{\ln(k)}{k} - \frac{\ln(k-1)}{k} \right)$. With infinite series, grouping terms or tossing in parentheses isn’t free—it can mess with the sum or convergence. Take $1 - 1 + 1 - 1 + \cdots$: leave it alone, it oscillates between 1 and 0, no convergence; group it like $(1 - 1) + (1 - 1) + \cdots$, it’s 0. Same series, different results. Your regrouping needs proof it matches the original, and you haven’t shown that.

The original $\sum_{k=1}^{\infty} \frac{\ln(k)}{k(k+1)}$ converges—compare it to $\frac{\ln(k)}{k^2}$, which works by integral test. But your split forms don’t hold up without tighter justification. Try partial sums or an integral trick instead.The original converges, sure—$\frac{\ln(k)}{k(k+1)} \approx \frac{\ln(k)}{k^2}$, and that works. But your splits need work. And what’s this Wolfram Alpha bit about $\gamma_n$? Did it spit out Stieltjes constants for partial sums? That’s a leap—maybe tied to $\zeta'(s)$, but you’d need to show how. Try partial sums or integrals instead of guessing.

If I am wrong, please downvote and I will delete it after receiving it.

Answer 2

Since we know the corresponding integral $$\int\frac{\log(k)}{k(k+1)}\,dk=\frac{1}{2}\log ^2(k)-\log (k+1) \log (k)-\text{Li}_2(-k)$$ $$\int_1^\infty\frac{\log(k)}{k(k+1)}\,dk=\frac{\pi ^2}{12}$$ we can use Euler-Maclaurin summation.

The summation will give $$S_n=\sum_{k=1}^n\frac{\log(k)}{k(k+1)}=\frac{\pi ^2}{12}+C_p+\sum_{m=1}^p (-1)^m \,\frac{a_m\,\log(n) +b_m}{a_m\,n^m }+O\left(\frac{\log(n)}{n^{p+1}}\right)$$ where $C_p$ is a small constant.

For example $$S_n=\frac{\pi ^2}{12}+C_3-\frac{\log (n)+1}{n}+\frac{4 \log (n)+1}{4 n^2}-\frac{36 \log (n)+1}{36 n^3}+O\left(\frac{\log(n)}{n^{4}}\right)$$

Either, you neglect $C_3$ or you rescale it using for example $$S_{100}=0.7329597935104060631852071\cdots$$