Proof that $\sum_{k=0}^\infty\frac{(-1)^k}{k+1}\sum_{m=0}^{\lfloor \frac{k}{2}\rfloor}( {\tiny\begin{matrix}2m\\m\end{matrix}})(-\frac14)^m$ converges Loosely connected to this problem, I want to show that the following infinite sum converges:
$$
\sum_{k=0}^\infty\frac{(-1)^k}{k+1}\sum_{m=0}^{\lfloor \frac{k}{2}\rfloor}\begin{pmatrix}2m\\m\end{pmatrix}\Big(-\frac14\Big)^m\tag{1}
$$
The hand-waving argument here is that $\sum_{m=0}^\infty ( {\scriptsize\begin{matrix}2m\\m\end{matrix}})(-\frac14)^m=\frac{1}{\sqrt{2}}$ so for large $k$ this factor is "approximately constant" and $\sum_{k=0}^\infty(-1)^k/(k+1)$ itself converges to $\log(2)$. However, I wasn't able to prove this so far. In the linked question, the answer of "GH from MO" provides some $\mathcal O$ argument to show that
$$
\lim_{n\to\infty}\sum_{m=0}^{\lfloor\frac{n-1}2\rfloor} \begin{pmatrix}2m\\m\end{pmatrix}\Big(-\frac14\Big)^m \sum_{k=2m}^{n-1}\frac{(-1)^k}{k+1}\tag{2}
$$
(note that (1) and (2) are equivalent problems as is readily verified) converges. Sadly I have very little experience with this notation so I'm not able to write this down in a detailed and rigorous way - hence why I want to find another / an easier way to prove this.

I tried to apply the Leibniz criterion but the sequence $(a_k)_{k\in\mathbb N_0}$ defined via  $a_k:=\frac1{k+1}\sum_{m=0}^{\lfloor k/2\rfloor} ( {\scriptsize\begin{matrix}2m\\m\end{matrix}})(-\frac14)^m$ is not monotonous (although positive and converges to 0). Also one could maybe use
$$
\begin{pmatrix}2m\\m\end{pmatrix}\sim 4^m/\sqrt{\pi m}\quad\text{ for }\quad m\to\infty
$$
but I wasn't able to apply this in a rigorous way to (1). I further considered summation by parts or the Cauchy criterion but at first glance those didn't seem too nicely applicable here.

Those are my efforts so far, I hope you'll be able to help me out on this one. Thanks in advance for any answer or comment!
 A: $$\sum_{m\geq 0}\frac{(-1)^m}{4^m}\binom{2m}{m}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta}=\frac{1}{\sqrt{2}} $$ and Dirichlet's test ensure that the given series is convergent. If $k$ is even (say $k=2n$) we have
$$ \sum_{m=0}^{n}\binom{2m}{m}\frac{(-1)^m}{4^m}=[x^n]\frac{1}{(1-x)\sqrt{1+x}} $$
and if $k$ is odd (say $k=2n+1$) we have the same identity, where $[x^n]f(x)$ stands for the coefficient of $x^n$ in the Maclaurin series of $f(x)$. In particular the original series can be written as
$$ \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)}\cdot[x^{2n}]\frac{1}{(1-x^2)\sqrt{1+x^2}}=\int_{0}^{1}\frac{1-x}{(1-x^2)\sqrt{1+x^2}}\,dx $$
since $\int_{0}^{1}x^{2n}(1-x)\,dx = \frac{1}{(2n+1)(2n+2)}$. It turns out that the original series is just
$$ \int_{0}^{1}\frac{dx}{(1+x)\sqrt{1+x^2}}=\frac{\text{arcsinh}(1)}{\sqrt{2}}=\color{red}{\frac{\log(1+\sqrt{2})}{\sqrt{2}}}. $$

I believe this has some meta-mathematical value too, in stressing how our brain does not really parse/learn equalities as symmetric objects. The very same argument read upside-down looks like a standard conversion of an integral into a series, while the conversion of the original series into a simple integral might appear particularly tricky at first sight. But such slickness is only apparent.
A: The series is 
$$\sum_{k=0}^{\infty}\frac{b_k}{k+1}\qquad\text{with $b_k=(-1)^k\sum_{m=0}^{\lfloor \frac{k}{2}\rfloor}\begin{pmatrix}2m\\m\end{pmatrix}\Big(-\frac14\Big)^m$}.$$
By Dirichlet's test, it suffices to show that the following sequence $\{B_n\}_n$ is bounded
$$B_n:=\sum_{k=0}^nb_k=
\sum_{m=0}^{\lfloor\frac{n}2\rfloor} \begin{pmatrix}2m\\m\end{pmatrix}\Big(-\frac14\Big)^m \sum_{k=2m}^{n}(-1)^k
=\frac{(-1)^n+1}{2}\sum_{m=0}^{\lfloor\frac{n}2\rfloor} \begin{pmatrix}2m\\m\end{pmatrix}\Big(-\frac14\Big)^m 
$$
which holds because $\sum_{m=0}^\infty \binom{2m}{m}(-\frac14)^m$ is convergent (it is equal to $=1/\sqrt{2}$).
A: Evaluation of the Sum
$$
\begin{align}
\sum_{k=0}^\infty\frac{(-1)^k}{k+1}\sum_{m=0}^{\lfloor k/2\rfloor}\binom{2m}{m}\left(-\frac14\right)^m
&=\sum_{k=0}^\infty\left(\frac1{2k+1}-\frac1{2k+2}\right)\sum_{m=0}^k\binom{2m}{m}\left(-\frac14\right)^m\tag1\\
&=\sum_{m=0}^\infty\binom{2m}{m}\left(-\frac14\right)^m\sum_{k=m}^\infty\left(\frac1{2k+1}-\frac1{2k+2}\right)\tag2\\
&=\sum_{m=0}^\infty\binom{2m}{m}\left(-\frac14\right)^m\int_0^1\frac{x^{2m}}{1+x}\,\mathrm{d}x\tag3\\
&=\int_0^1\frac{\mathrm{d}x}{(1+x)\sqrt{1+x^2}}\,\tag4\\
&=\int_0^{\pi/4}\frac{\mathrm{d}u}{\sin(u)+\cos(u)}\,\tag5\\
&=\int_0^{\pi/4}\frac{\mathrm{d}u}{\sqrt2\cos(u)}\,\tag6\\[3pt]
&=\frac1{\sqrt2}\log(\sqrt2+1)\tag7
\end{align}
$$
Explanation:
$(1)$: separate even $k$ as $2k$ and odd $k$ as $2k+1$
$(2)$: switch order of summation
$(3)$: write sum as an integral (see $(13)$ from this answer for acceleration)
$(4)$: sum in $m$ using the Taylor series for $(1+x)^{-1/2}$
$(5)$: substitute $x=\tan(u)$
$(6)$: substitute $u\mapsto\pi/4-u$
$(7)$: $\int\sec(u)\,\mathrm{d}u=\log(\sec(u)+\tan(u))+C$ 
Above, we used the Taylor series
$$
(1+x)^{-1/2}=\sum_{m=0}^\infty\binom{2m}{m}\left(-\frac x4\right)^m\tag8
$$

Shortcut
We can deduce convergence at $(1)$ because
$$
0\le\left(\frac1{2k+1}-\frac1{2k+2}\right)\le\frac1{(2k+1)^2}\tag9
$$
and because $\binom{2m}{m}\frac1{4^m}=\frac{4m-2}{4m}\binom{2m-2}{m-1}\frac1{4^{m-1}}\le\binom{2m-2}{m-1}\frac1{4^{m-1}}$ shows that
$$
\sum\limits_{m=0}^\infty\binom{2m}{m}\left(-\frac14\right)^m\tag{10}
$$
converges by the alternating series test, and the partial sums of $(10)$ are between $1$ and $\frac12$.
