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

Question

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.

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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.

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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!

Answer 1

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 $=1/\sqrt{2}$).

Answer 2

$$\sum_{m\geq 0}\frac{(-1)^m}{4^m}\binom{2m}{m}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta}=\frac{1}{\sqrt{2}} $$ 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 $$ \int_{0}^{1}\frac{dx}{(1+x)\sqrt{1+x^2}}=\frac{\text{arcsinh}(1)}{\sqrt{2}}=\color{red}{\frac{\log(1+\sqrt{2})}{\sqrt{2}}}. $$

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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.

Answer 3

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 $$ (1+x)^{-1/2}=\sum_{m=0}^\infty\binom{2m}{m}\left(-\frac x4\right)^m\tag8 $$

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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 $\frac12$.