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.
 A: $$\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}=\sum_{k}\left(\frac{\ln(4k-3)}{4k-3}-\frac{\ln(4k-1)}{4k-1} \right)[k \ge 1].$$
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.
A: 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.
A: 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: $\sum_{k=1}^{\infty} (a_{k+1}-a_k)= (\lim_{k\to\infty}a_{k+1})- a_1$.
