I'm looking for series alike in the literature containing in their summands numbers of the type $\displaystyle Q_n=\sum _{k=0}^{n-1}\frac{1}{2^k }\binom{2 k}{k}$. The following three series have recently been presented by C.I.Valean in this short exposition,
\begin{equation*}i) \ \sum _{n=1}^{\infty } \frac{ 2^n}{\displaystyle n^2 \binom{2 n}{n}}\sum _{k=0}^{n-1}\frac{1}{2^k }\binom{2 k}{k}=\sum _{n=1}^{\infty } \frac{ 2^n Q_n}{\displaystyle n^2 \binom{2 n}{n}}=2 G;\end{equation*}\begin{equation*}\small ii) \ \sum _{n=1}^{\infty } (-1)^{n-1} \frac{ 2^n}{\displaystyle n^2 \binom{2 n}{n}}\sum _{k=0}^{n-1}\frac{1}{2^k }\binom{2 k}{k}=\sum _{n=1}^{\infty } (-1)^{n-1} \frac{ 2^n Q_n}{\displaystyle n^2 \binom{2 n}{n}}=\frac{\pi}{3} \log(2+\sqrt{3})-\frac{2}{3}G;\end{equation*}\begin{equation*}iii) \ \sum _{n=1}^{\infty } \frac{ 4^n}{\displaystyle n^2 \binom{4 n}{2n}}\sum _{k=0}^{2n-1}\frac{1}{2^k }\binom{2 k}{k}=\sum _{n=1}^{\infty } \frac{ 4^n Q_{2n}}{\displaystyle n^2 \binom{4 n}{2n}}=\frac{16}{3}G-\frac{2}{3}\pi \log(2+\sqrt{3}),\end{equation*}where here $\displaystyle G=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)^2}$ is the Catalan's constant.
For example, in the case of the second series, we may proceed as follows: exploiting that $\displaystyle \int_0^{\pi/2} \sin(n \theta) \cos^{n-1}(\theta) \textrm{d}\theta=\frac{2^n}{\displaystyle n \binom{2n}{n}}\sum_{k=0}^{n-1} \frac{1}{2^k}\binom{2k}{k}$, which is found and evaluated in More (Almost) Impossible Integrals, Sums, and Series (2023), page $16$ (exploiting simple recurrence relations is enough to extract a solution), we have$$\sum _{n=1}^{\infty } (-1)^{n-1} \frac{ 2^n}{\displaystyle n^2 \binom{2 n}{n}}\sum _{k=0}^{n-1}\frac{1}{2^k }\binom{2 k}{k}=\sum _{n=1}^{\infty} (-1)^{n-1}\frac{1}{n} \int_0^{\pi/2} \sin(n \theta) \cos^{n-1}(\theta) \textrm{d}\theta$$$$=\int_0^{\pi/2}\frac{1}{\cos(\theta)}\Im \biggr\{\sum_{n=1}^{\infty} (-1)^{n-1}\frac{(\cos(\theta) e^{i \theta})^n}{n}\biggr\}\textrm{d}\theta=\int_0^{\pi/2} \frac{\Im \{\log(1+\cos(\theta) e^{i \theta})\}}{\cos(\theta)}\textrm{d}\theta$$
$$\small =\int_0^{\pi/2}\frac{1}{\cos(\theta)} \arctan\left(\frac{\cot(\theta)}{1+2\cot^2(\theta)}\right)\textrm{d}\theta\overset{\theta\mapsto \pi/2-\theta}{=}\int_0^{\pi/2}\frac{1}{\sin(\theta)} \arctan\left(\frac{\tan(\theta)}{1+2\tan^2(\theta)}\right)\textrm{d}\theta$$$$\overset{\theta \mapsto 2 \arctan(\theta)}{=}\int_0^1 \frac{1}{\theta}\arctan\left(\frac{2\theta (1-\theta^2)}{1+6 \theta^2+\theta^4}\right)\textrm{d}\theta$$$$=\int_0^1 \frac{1}{\theta}\arctan\left(\frac{4\theta/(1-\theta^2)-2\theta/(1-\theta^2)}{1+(4 \theta/(1-\theta^2)) \cdot (2 \theta/(1-\theta^2))}\right)\textrm{d}\theta$$$$=\int_0^1 \frac{1}{\theta}\arctan\left(\frac{4\theta}{1-\theta^2}\right)\textrm{d}\theta-\int_0^1 \frac{1}{\theta}\arctan\left(\frac{2\theta}{1-\theta^2}\right)\textrm{d}\theta$$$$=\int_0^1 \frac{1}{\theta}\arctan\left(\frac{(2-\sqrt{3})\theta+(2+\sqrt{3})\theta}{1-(2-\sqrt{3})\theta\cdot (2+\sqrt{3})\theta}\right)\textrm{d}\theta-2\int_0^1 \frac{\arctan(\theta)}{\theta}\textrm{d}\theta$$$$=\int_0^1 \frac{\arctan((2-\sqrt{3})\theta)}{\theta}\textrm{d}\theta+\int_0^1 \frac{\arctan((2+\sqrt{3})\theta)}{\theta}\textrm{d}\theta-2\int_0^1 \frac{\arctan(\theta)}{\theta}\textrm{d}\theta$$$$=\operatorname{Ti}_2(2-\sqrt{3}) +\operatorname{Ti}_2(2+\sqrt{3}) -2 \operatorname{Ti}_2(1)=\frac{\pi}{3} \log(2+\sqrt{3})-\frac{2}{3}G,$$where we also needed to employ the well-known special values of the Inverse tangent integral like $\displaystyle \operatorname{Ti}_2(1)=G$, $\displaystyle \operatorname{Ti}_2(2-\sqrt{3})=\frac{2}{3}G+\log(2-\sqrt{3})\frac{\pi}{12}$, and $\displaystyle \operatorname{Ti}_2(2+\sqrt{3})=\frac{2 }{3}G-\frac{5}{12}\log \left(2-\sqrt{3}\right)\pi$.
For another solution, one might think of using Wallis' integrals.
Question 1: Given the simplicity of the first result I wonder if that series already appeared in the literature (if so, the presentation will be updated accordingly). Have you ever met it before (together with a reference)?
Question 2: More generally, I'm interested in finding more series involving those $Q_n$ numbers if they are available in the literature. Any references?
Personally, I would also find very interesting more general cases like $$S_m=\sum _{n=1}^{\infty } \frac{ 2^n Q_n}{\displaystyle n^m \binom{2 n}{n}}, \ m \ge3,$$and note that for $m=2$ we get the series from the point $i)$.
Update 1:
Another fantastic example, given the generalization above, is the case $m=3$. Cornel says that by using a strategy similar to the one in the reference above, we get that$$\sum _{n=1}^{\infty } \frac{ 2^n}{\displaystyle n^3 \binom{2 n}{n}}\sum _{k=0}^{n-1}\frac{1}{2^k }\binom{2 k}{k}=\sum _{n=1}^{\infty } \frac{ 2^n Q_n}{\displaystyle n^3 \binom{2 n}{n}}=2 \log(2)G,$$and it looks too beautiful to be real!
This closed form makes me wonder if (there is any chance) we can relate this series to this integral problem in this post How to show $\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G$.
Update 2: Okay, I've just found an answer to my thoughts in the previous update, and, Yes, they can be perfectly related. So, we can, for example, evaluate this series at this point by three solutions (a bit subtle this part in the sense that they are different only from the perspective of getting the core integral evaluated differently), and now I refer to the integral at the reference in the previous update (it represents the toughest obstacle in the evaluation). So, for that integral we already have two solutions in More (Almost) Impossible Integrals, Sums, and Series (2023) and one by Sujeethan Balendran available at that link.
Update 3: The version $m=4$ has just been derived by Cornel, and together with the closed form all looks amazing,
$$\sum _{n=1}^{\infty } \frac{ 2^n}{\displaystyle n^4 \binom{2 n}{n}}\sum _{k=0}^{n-1}\frac{1}{2^k }\binom{2 k}{k}=\sum _{n=1}^{\infty } \frac{ 2^n Q_n}{\displaystyle n^4 \binom{2 n}{n}}$$$$=\log^3(2)\frac{\pi}{12}-\frac{\pi ^2 }{3}G-\frac{5 }{48}\pi ^4 +\frac{5 }{384}\psi ^{(3)}\left(\frac{1}{4}\right)$$$$-4 \log (2) \Im\left \{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-8 \Im\left \{\text{Li}_4\left(\frac{1+i}{2}\right)\right\}.$$
In the derivation process, the same starting ideas in the paper above were exploited.