My question concerns the (generalized) Legendre polynomials in the $d$-dimensional space (see, e.g. Mueller, Freeden & Gutting): $$ P_n(d;t) = \frac{|S^{d-3}|}{|S^{d-2}|} \int_{-1}^1 (t + i \sqrt{1-t^2}s)^n\ (1-s^2)^{\frac{d-4}{2}}ds $$ where $|S^{k-1}|$ denotes the area of the unit sphere in $\mathbb{R}^k$. When $d=3$, $P_n(d;t)$ is the regular Legendre polynomial. For general $d$, $P_n(d;t)$ is closely related to the spherical harmonics on $S^{d-1}$.
Using the integral representation, it is a standard result that $$ |P_n(d;t)| \leq \frac{\Gamma(\frac{d-1}{2})}{\sqrt\pi}\left(\frac{4}{n(1-t^2)}\right)^{\frac{d-2}{2}}, \quad |t|<1. $$
Also using the integral representation, it is easy to obtain that $$ \left| \sum_{n=m}^{m+p} (-1)^n P_{2n}(d;t) \right| \leq \frac{C_d}{|t| \cdot (n(1-t^2))^{\frac{d-2}{2}}} $$ for arbitrary $m,p > 0$, where $C_d$ is a constant that only depends on $d$. Using the integral representation, this just follows from bounding the sum of the geometric progression $\sum_n (-1)^n (t+i\sqrt{1-t^2s})^{2n}$ inside the integrand.
Question: Numerical simulation seems to suggest that $$ \left| \sum_{n=m}^{m+p} (-1)^n n^{\frac{d-2}{2}} P_{2n}(d;t) \right| $$ is also uniformly bounded for all $m,p>0$. But now one cannot simply bring the sum under the integral because the sum may not have a bounded norm. (I also tried a few other integral representations but none seems to work with this idea.) I am wondering if there is a different argument that can show the sum above is uniformly bounded. Thanks!