Is there a closed form for the series $\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}?$

This is actually a question which was deleted by a previous user. I worked very hard on this question and became quite engrossed. The question was simply: Is there a closed form for the series $$\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}?$$

I am submitting my work below as an answer and I'd like Math.SE to help me out here. There is likely something I've missed, and I would like some help on figuring out if there is a closed form.

Solution 1:

We can rewrite the given sum as $$ \begin{align*} - \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\log(2n+1)}{2n+1} &= - \left. \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\log(2n+1)}{(2n+1)^s} \right|_{s=1} \\ &= - \left. \frac{d}{ds} \right|_{s=1} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^s} \\ &= - \beta'(1) \\ &= - \frac{\pi}{4} \Bigl\{\gamma + 2 \log 2 + 3 \log \pi - 4 \log\left[\Gamma(1/4)\right]\Bigr\}, \end{align*} $$ where $\beta$ is the Dirichlet beta function. The last equality was found at MathWorld, which cites an OEIS entry (among other things), which in turn cites the MathWorld article.

Solution 2:

While not conclusive proof that there doesn't exist a closed form for this sum, this indicates that there is no simple closed form. The value computed might also act as a hint to the (in)validaty of a closed form.

This series converges very slowly. Using the Euler-Maclaurin Sum Formula, we get the asymptotic expansion:

$$ \begin{array}{l} \sum_{k=1}^n\frac{\log(4k-a)}{4k-a}\\ \sim\small\left(\frac1{3(4n{-}a)^2}-\frac{44}{45(4n{-}a)^4}+\frac{8768}{945(4n{-}a)^6}-\frac{30976}{175(4n{-}a)^8}+\frac{58400768}{10395(4n{-}a)^{10}}\right)\\ \small+\left(\frac{1}{2(4n{-}a)}-\frac{1}{3 (4n{-}a)^2}+\frac{8}{15 (4n{-}a)^4}-\frac{256}{63 (4n{-}a)^6}+\frac{1024}{15 (4n{-}a)^8}-\frac{65536}{33 (4n{-}a)^{10}}\right) \log (4n{-}a)\\ \small+\frac{1}{8} \log ^2(4n{-}a)+C_a \end{array} $$ Note that all the terms vanish as $n\to\infty$ except the last two. Furthermore, $$ \begin{align} 0 &\le\lim_{n\to\infty}\left(\log^2(4n-1)-\log^2(4n-3)\right)\\[9pt] &=\lim_{n\to\infty}\left(\log(4n-1)+\log(4n-3)\right)\left(\log(4n-1)-\log(4n-3)\right)\\ &\le\lim_{n\to\infty}2\log(4n-1)\log\left(1+\frac2{4n-3}\right)\\ &\le\lim_{n\to\infty}\frac{4\log(4n-1)}{4n-3}\\[6pt] &=0 \end{align} $$ Therefore, $$ \sum_{k=1}^\infty\frac{\log(4k-3)}{4k-3}-\frac{\log(4k-1)}{4k-1}=C_3-C_1 $$ Plugging in $n=10000$ into the asymptotic expansion for $a=1$ and $a=3$, we get to $50$ places $$ \begin{align} C_1&=+0.03827947109192929052721132046406192021405884328730\\ C_3&=-0.15462184570498313883597844356397086503103802433270\\ C_3-C_1&=-0.19290131679691242936318976402803278524509686762001 \end{align} $$ Using the Inverse Symbolic Calculator, there doesn't seem to be a simple expression for this number.

Solution 3:

$$\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}=\sum_{k}\left(\frac{\ln(4k-3)}{4k-3}-\frac{\ln(4k-1)}{4k-1} \right)[k \ge 1].$$

Let $k=j+1$. As a result, $$4k-3=4j+4-3=4j+1, 4k-1=4j+4-1=4j+3, \text{ and } k\ge 1 \implies j \ge 0.$$

$$\sum_{k}\left(\frac{\ln(4k-3)}{4k-3}-\frac{\ln(4k-1)}{4k-1} \right)[k \ge 1]=\sum_j \left( \frac{\ln(4j+1)}{4j+1}-\frac{\ln(4j+3)}{4j+3}\right)[j \ge 0].$$

There is no 'nice' cancelling that occurs between these two summands. This is because the set $A=\{4j+1: j\ge 0\}$ and $B=\{4j+3: j \ge 0\}$ have no elements of intersection; that is, $A\cap B=\emptyset.$ While it isn't entirely relevant, this can be formally proven by noting $A$ and $B$ are equivalence classes $[1]=\{b \in \mathbb{Z}:b\equiv 1 \pmod 4\}$ and $[3]=\{b \in \mathbb{Z}: b\equiv 3\pmod 4\}$, respectively. $[1]$ and $[3]$ partition $\mathbb{Z}$ into disjoint sets, hence $A\cap B=\emptyset$. That begs the question: Where to from here?

To be honest, I don't know. I cannot conclusively prove that there is not a closed form solution for this series, but I certainly cannot find it. It does converge, and the integral test provides a nice upperbound for it.

I hope someone else can be of assistance.

Solution 4:

Hint. For all sequence $\{b_k\}_{k\in\mathbb{N}}$ we have \begin{equation} \begin{split} \sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)} & = \sum_{k=1}^\infty \left[\frac{\ln(4k-3)}{(4k-3)}+ b_k\right] - \left[ \frac{\ln(4k-1)}{(4k-1)} + b_k\right] \\ \end{split} \end{equation} Supose that $\sum_{k=1}^\infty b_k$ converge and
$$ a_{k+1}= \left[ \frac{\ln(4k-3)}{(4k-3)}+b_k\right] \quad a_{k}= \left[\frac{\ln(4k-1)}{(4k-1)}+ b_k\right] $$ Use the telescopic propert of series: $\sum_{k=1}^{\infty} (a_{k+1}-a_k)= (\lim_{k\to\infty}a_{k+1})- a_1$.