A need for analytic continuation of a finite sum function

Question

Let $\varphi(n):=(-1)^{n+1}(n+1)2^{2n}$.

I am able to prove the following identity (${\color{red}{\mathbf{LHS}}}$=infinite series, ${\color{blue}{\mathbf{RHS}}}$=finite sum) \begin{align*} {\color{red}{\frac{1}{\varphi(n)}\sum_{k\geq1}\frac{(-1)^k}{\binom{3n+k+2}{2n+2}}}} &={\color{blue}{\sum_{j=1}^n\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}+3-4\ln 2}}. \tag1 \end{align*} Replacing $n=0$ in equation (1) leads to $\displaystyle\sum_{j\geq1}\frac{(-1)^j}{\binom{j+2}2}=4\ln 2-3$.

Letting $n\rightarrow\infty$ in (1) unearths the less trivial result $$\sum_{j=1}^{\infty}\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}=4\ln 2-3.$$

QUESTION. How can one exercise an analytic continuation of the ${\color{blue}{\mathbf{RHS}}}$ in (1), ${\color{purple}{\text{in the variable $n$}}}$, so that we would be able to compute at $n=-\frac23$ and derive (from it) a value for the infinite series on the ${\color{red}{\mathbf{LHS}}}$, i.e. $$\frac1{\varphi(-\frac23)}\sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}\,?$$

Answer 1

As for the sum $$\sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}$$ one can evaluate it by means of the Beta function integral, like in this recent computation. $$\sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}=\frac23\sum_{k\geq1}(-1)^k \frac{\Gamma\big(k+\frac13\Big)\Gamma\big(\frac23\big)}{\Gamma(k+1)}=\frac23\sum_{k\geq1}(-1)^kB\Big(k+\frac13,\frac23\Big)=$$ $$=\frac23\sum_{k\geq1} \int_0^1(-1)^k {x^{k-\frac23}}{(1-x)^{-\frac13}}dx=-\frac23 \int_0^1\frac{1}{1+x} \Big(\frac{x }{1-x}\Big)^{1/3}dx=$$

The latter integral is easily rationalised by changing variable; we get $$=-\int_0^\infty\frac{2t^3}{(t^3+1)(2t^3+1)}dt=\int_0^\infty\frac{2dt}{2t^3+1 } -\int_0^\infty\frac{2dt}{ t^3+1}=$$$$=(\sqrt[3]4-2)\int_0^\infty\frac{dt}{ t^3+1}$$ whence $$\sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}= \frac{2\pi}{3\sqrt3} (\sqrt[3]4-2)$$ which evaluates to $-0.4989..$.

Answer 2

\begin{align*} \sum_{k\geq 1} (-1)^k \frac{1}{\binom{k}{x}} &=x!(-x)!\sum_{k\geq1}(-1)^k\frac{(-x+k)!}{(-x)!\,k!} =x!(-x)!\sum_{k\geq1}\binom{x-1}{k} \\ &=x!(-x)!(2^{x-1}-1)=\frac{\pi x}{\sin(\pi x)} (2^{x-1}-1):=f(x) \end{align*} for $0< x < 2$, straightforwardly from $(-1)^k \binom{\alpha-1+k}{k}=\binom{-\alpha}{k}$ with $\alpha-1=-x$.

Hence $f(2/3)\simeq -.498914$

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Remarks:

(I'm way behind in organizing and formatting many sets of notes into semi-coherent pdfs, so I'm too pressed for time to pursue this to any further depth, but I'm tempted to test (extend my understanding of) some umbral Sheffer operational calculus on this though.)

An infinite sum resolved to a finite sum is typically what you get with a generalized Dobinski-type formula or Euler binomial transform of an e.g.f. where one side has coefficients polynomial in $k$ (see, e.g., this MO-Q on the Dobinski formula).

With $B_n(x)$ the Bell / Touchard / Stirling polynomials of the second kind,

$$g(B.(x)) = e^{-x} \sum_{k \geq 0} g(k) \frac{x^k}{k!} =e^{-x}e^{g.x} = e^{-(1-g.)x}$$

$$ = \sum_{k \geq 0} (-1)^k(1-g.)^k \frac{x^k}{k!} = \sum_{k \geq 0} (-1)^k\triangledown_{j=0}^k g(j) \frac{x^k}{k!}$$

and the last is a finite sum when $g(j)$ is a polynomial. Multiplying by $(1+x)$ sums up pairs of coefficients.

Umbrally substitute the falling factorial polynomial, a.k.a. the Stirling polynomials of the first kind, $(x)_m = m!\binom{x}{m}$ for $x$ in $g(B.(x))$ to retrieve $g(x)$ as the analytic continuation of $g(n)$.

I wonder if you can approach the problem this way to relate the LHS to the RHS, eventually evaluating at $x=1$.

Answer 3

I liked the two contributions by Pietro Majer. One is the evaluation of the series in his answer box, the other is a pragmatic direction towards extending the sum for real values of $n$. I would like to expand on the latter.

Denoting the function $$S(N):={\color{blue}{\sum_{j=1}^N\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}}}$$ one can perform the analytic continuation by the following procedure \begin{align*} S(N)&=\sum_{j=1}^{\infty}-\sum_{j=N+1}^{\infty} \\ &=4\ln 2-3- \sum_{j=1}^{\infty}\frac{(-1)^j(28(j+N)^2+24(j+N)+4)}{\binom{3j+3N}{j+N}(j+N) (3j+3N+2)(3j+3N+1)\,2^{2j+2N}}. \end{align*} Therefore, \begin{align*} S(-2/3)&=4\ln 2 -3-\sum_{j=1}^{\infty}\frac{(-1)^j(252j^2-120j+4)}{9\binom{3j-2}{j-\frac23}\,j\,(3j-2) (3j-1)\,2^{2j-\frac43}} \end{align*} and hence \begin{align*} \sum_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1} &=\sum_{j=1}^{\infty}(-1)^j\frac{252j^2-120j+4}{27j\,(3j-2) (3j-1)\,2^{2j}}\binom{3j-2}{j-\frac23}^{-1}. \end{align*} Ripping the benefit: we have found a "closed form" for a fairly complicated infinite series $$\sum_{j=1}^{\infty}(-1)^j\frac{252j^2-120j+4}{27j\,(3j-2) (3j-1)\,2^{2j}}\binom{3j-2}{j-\frac23}^{-1} =\frac{2\pi}{3\sqrt3} (\sqrt[3]4-2).$$

Another possible approach for an analytic continuation is: let $f\left(x\right)\in C^{1}\left[a,b\right]$, then \begin{align*} \sum_{a<n\leq b}f\left(n\right)&=\int_{a}^{b}f\left(x\right)dx+\int_{a}^{b}\left(x-\left\lfloor x\right\rfloor-\frac{1}{2}\right)f'\left(x\right)dx \\ & \qquad \qquad \qquad +\left(a-\left\lfloor a\right\rfloor-\frac{1}{2}\right)f\left(a\right)-\left(b-\left\lfloor b\right\rfloor-\frac{1}{2}\right)f\left(b\right) \end{align*} for suitable choices of $a$ and $b$. Here, $\lfloor\cdot\rfloor$ stands for the floor function.