Prove or disprove $\sum_{i=0}^\infty \frac{1}{i+j+1}\frac{1}{\sqrt{i+\frac{1}{2}}}<\frac{\pi}{\sqrt{j+1}}$ In Example 2.3.5; Functional Analysis book by S. Kesavan it was shown that for $j\gt -\frac{1}{2}$
$$s(j) := \sum_{i=0}^\infty \frac{1}{i+j+1}\frac{1}{\sqrt{i+\frac{1}{2}}}<\frac{\pi}{\sqrt{j+\frac{1}{2}}}\tag{1}$$
Numerically, even the stronger inequality seems to hold
$$s(j) <\frac{\pi}{\sqrt{j+1}}\tag{2}$$
I have proved $(1)$ by comparing the sum with an integral, but I didn't succeed to prove $(2)$. Can you find a proof?

In S. Kesavan's example, we do no need the summation to begin from $i=0$. As $a_{ij}$ is not even defined there. Therefore, I was able to prove a relaxed version of the inequality.
$$\sum_{i=1}^{\infty}\frac{1}{(i+j+1)\sqrt{i+1/2}} = \frac{1}{(j+2)\sqrt{3/2}}+\sum_{i=2}^{\infty}\frac{1}{(i+j+1)\sqrt{i+1/2}} \tag{1}$$
Further,
$$\sum_{i=2}^{\infty}\frac{1}{(i+j+1)\sqrt{i+1/2}} < \sum_{i=2}^{\infty}\frac{1}{(i+j+1)\sqrt{i}}<\sum_{i=2}^{\infty}(\int_{i-1}^{i}\frac{dx}{(x+j+1)\sqrt{x}})$$
$$\sum_{i=2}^{\infty}(\int_{i-1}^{i}\frac{dx}{(x+j+1)\sqrt{x}})=\int_{1}^{\infty}\frac{dx}{(x+j+1)\sqrt{x}} \tag{2}$$
Also,
$$\frac{1}{(j+2)\sqrt{3/2}}<\frac{1}{(j+2)}<\int_{0}^{1}\frac{dx}{(x+j+1)\sqrt{x}} \tag{3}$$
From $(1),(2)$ and $(3)$ we have,
$$ \sum_{i=1}^{\infty}\frac{1}{(i+j+1)\sqrt{i+1/2}} < \int_{0}^{\infty}\frac{dx}{(x+j+1)\sqrt{x}} = \frac{\pi}{\sqrt{j+1}} $$
