# binomial transform, Hurwitz zeta function

For $j,n\in\mathbb Z_+$, let $$L_{j,n}^{(t)}= \sum_{m=0}^{n} \Bigl(-\frac 12\Bigr)^{n-m}{n\choose m}{m+j+1\choose m+1} \left( \frac {1}{t+\frac 12}\right)^{m+j+2}$$ and $$L_{j,n} =\sum_{t=1}^\infty L_{j,n}^{(t)}$$

We would like to show that for $j,n\ge 0$ $$(*)\quad \quad L_{2j,2n}>0.$$ Hypothesis (*) is supported by the numerics we did. In fact, it seems that $L_{j,n}>0$ is positive for all $j>0$ and all $n$. However, we only need the result for even indexes.

What do we know? Namely, the following easy facts:

• $L_{j,0}>0$ for all $j$.
• $L_{0,2n}>0$ for all $n$.
• $L_{j,n}^{(1)}>0$ for all $n$ and $j>0$.

These easy facts follow from the observation that

$$L_{j,n}^{(t)}=\frac {(-1)^j}{j!}\frac {d^j}{dz^j} \left( \frac {1}{z^2} \left[ \frac {1}{z}-\frac 12\right]^n \right)\left|_{z=t+\frac 12}\right.$$

In general, we need to estimate the sign of $$\sum_{t=1}^\infty \Phi_n^{(j)}(t+\frac 12),\quad\text{where } \Phi_n(z)=\frac {1}{z^2} \Bigl[ \frac {1}{z}-\frac 12\Bigr]^n.$$ Heuristically, $$(-1)^j j! L_{j,n}=\sum_{t=1}^\infty \Phi_n^{(j)}(t+\frac 12) \approx -\Phi_n^{(j-1)}(\frac 32)$$ and the sign of $\Phi_n^{(j)}(z)$ is $(-1)^j$ for every $z<2$, which supports the conjecture as well.

-