Closed form of $\sum_{k=0}^n\binom{2k}{k}(-1/4)^k$

Question

Loosely related to my last question, I was trying to find a closed form of the finite sum

$$a_n:=\sum_{k=0}^n\binom{2k}{k}\left(-\frac{1}4\right)^k$$

This is not too different from the well-known expression

$$\sum_{k=0}^n\binom{2k}{k}\left(\frac{1}4\right)^k=\binom{n+\frac12}{n}=\frac{2n+1}{2^{2n}}\begin{pmatrix}2n\\n\end{pmatrix}$$

so

$$ \sum_{k=0}^n\binom{2k}{k}\Big(\frac{-1}4\Big)^k=\sum_{k=0}^n\binom{2k}{k}\Big(\frac{1}4\Big)^k-2\sum_{\substack{k=0\\k\text{ odd}}}^n\binom{2k}{k}\Big(\frac{-1}4\Big)^k\\ =\frac{2n+1}{2^{2n}}\binom{2n}{n}-\frac12\sum_{l=0}^{\lfloor \frac{n-1}2 \rfloor}\begin{pmatrix}4l+2\\2l+1\end{pmatrix}\frac1{16^l}. $$

However, the second sum does not seem to be easier to handle at first glance. Also trying a similar ansatz $a_n= \binom{2k}{k}p_n/2^{2n}$ led to seemingly unstructured $p_n$ given by (starting from $p=0$):

$$ 1,1,\frac73,\frac95,\frac{72}{35},\frac{9}{7},\frac{185}{77},\frac{227}{143},\frac{5777}{2145},\ldots $$

So I was wondering (a) if there already is a known closed form for the $a_n$ (I've searched quite a bit but sadly wasn't successful) and if not, then (b) what is a good way to tackle this problem - or does a "simple" (sum-free) form maybe not even exist?

Thanks in advance for any answer or comment!

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Edit: I am aware of the solution

$$ a_n=(-1)^n 2^{-2n+2}\begin{pmatrix}2n+2\\n+1\end{pmatrix} {}_2F_1(1;n+\frac32;n+2;-1)+\frac1{\sqrt2} $$

Mathematica presents where ${}_2F_1$ is the hypergeometric function. But as the latter is given by an infinite sum, I was hoping for something simpler by connecting it to the known, non-alternating problem as described above - although I'd also accept if this was not possible.

Answer 1

$$\frac{(-1)^k}{4^k}\binom{2k}{k}=[x^k]\frac{1}{\sqrt{1+x}}$$ implies that $$ \sum_{k=0}^{n}\frac{(-1)^k}{4^k}\binom{2k}{k} = [x^n]\frac{1}{(1-x)\sqrt{1+x}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}[x^n]\frac{1}{1+x+\sqrt{2+2x}} $$ hence $a_n$ is positive and convergent to $\frac{1}{\sqrt{2}}$, and the hypergeometric representation is straightforward.
The Maclaurin coefficients of $g(x)=\frac{1}{(1-x)+\sqrt{2}\sqrt{1-x}}$ are positive and they behave like $\frac{1}{\sqrt{2\pi n}}$ for large values of $n$, hence we have the approximate identity $$ a_n \approx \frac{1}{\sqrt{2}}+\frac{(-1)^n}{2\sqrt{\pi n}} $$ for $n\to +\infty$.

Answer 2

I think that \begin{align} a_n&=\sum_{k=0}^n\binom{2k}{k}\biggl(-\frac14\biggr)^k\\ &=\frac2\pi\int_0^{\pi/2}\frac{(-1)^n \sin ^{2 n+2}x+1}{\sin ^2x+1}\textrm{d}x\\ &=\frac2\pi\int_0^{\infty}\frac{1}{2+x^2}\biggl[1+\frac{(-1)^n}{(1+x^2)^{n+1}}\biggr]\textrm{d}x. \end{align} These integral representations imply that

1. the sequence $a_n>0$ for all $n\in\{0\}\cup\mathbb{N}$;
2. the sequence $a_{2n}$ for all $n\in\{0\}\cup\mathbb{N}$ is decreasing and convex;
3. the sequence $a_{2n+1}$ for all $n\in\{0\}\cup\mathbb{N}$ is increasing and concave;
4. the limit $\lim_{n\to\infty}a_n=\frac{\sqrt{2}\,}{2}$ is valid.

If this formula is useful, please read Theorem 23 and its proof in the paper

Feng Qi and Bai-Ni Guo, Integral representations of the Catalan numbers and their applications, Mathematics 5 (2017), no. 3, Article 40, 31 pages; available online at https://doi.org/10.3390/math5030040.

I will write down the proof of my own idea above in a paper and then come back.

By the way, What is the general formula of the sum $\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{k/2}{m}$ for $m,n\in\mathbb{N}$?