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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!


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.

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    $\begingroup$ Please see OEIS sequences A003148 and related A091520. $\endgroup$ – Somos Apr 20 '18 at 17:19
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$$\frac{(-1)^k}{4^k}\binom{2k}{k}=[x^k]\frac{1}{\sqrt{1+x}}$$ 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 $n\to +\infty$.

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  • $\begingroup$ +1 Very nice. Although I'll take it there is no "nice" closed form like for $\sum_{k=0}^n\binom{2k}{k}(1/4)^k$, the asymptotic behaviour surely is very interesting! $\endgroup$ – Frederik vom Ende Apr 21 '18 at 16:15
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    $\begingroup$ @FrederikvomEnde: agreed - the singularities of $\frac{1}{1-x}$ and $\frac{1}{\sqrt{1+x}}$ do not interact nicely, but that is not an actual issue for the asymptotic study. $\endgroup$ – Jack D'Aurizio Apr 21 '18 at 16:31

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