Convergence of integral and summation for Time taken for complete revolution around vertical circle I want to find Time taken to complete Vertical circular motion by Particle of mass $m$
So I proceed as follows
Applying work energy theorem
$$-mgR(1-\cos(\theta)=\frac12 mv^2-\frac12 mu^2$$
$$\implies v^2=u^2-2gR(1-\cos\theta)$$
$$\implies \omega=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$
$$\implies \frac {d\theta}{dt}=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$
$$\implies \frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\frac{dt}{R}$$
Let T be the time taken for complete revolution
Now, Integrating both sides
$$\int_{0}^{2\pi}\frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\int_{0}^{T}\frac{dt}{R}$$
$$\implies T=2R\int_{0}^{\pi}\frac{d\theta}{\sqrt{u^2-4gR\sin^2\theta}}$$
$$\implies T=\frac{2R}{u}\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\frac{4gR}{u^2}\sin^2\theta}}$$
On evaluating this elliptic integral we get,
$$T=\sum_{n=0}^{\infty} \left(\frac{2}{u}\right)^{2n+1}\left(\frac{(2n-1)!!}{2^n n!}\right)^2 (gR)^n$$
Here $T$ denotes Time taken for complete revolution. Block will complete full circle only if $u\geq \sqrt{5gR}$. So my question is $T$ will be Real only when $u\geq \sqrt{5gR}$ so This condition should exist in integral as well as summation for $T$ but I'm not able to see it. Also Wolfram alpha evaluates integral only if $u\geq \sqrt{5gR}$

Both the integral and the series converge if and only if $u > 2 \sqrt{g R}$, but your formula for $T$ is correct only if $u \geq \sqrt{5 g R}$. This discrepancy can be understood as follows:
The first condition only ensures that the particle has enough energy to reach the height $2R$ (it can be found by equating $2 R m g$ and $\frac{1}{2} m u^2$).
But your formula for $T$ can only be correct if the particle slides along the loop at all times. This only happens if the sum of the centrifugal force and the radial component of the gravitational force is always non-negative, i.e. $m R \omega^2 + m g \cos (\theta) \geq 0$. Using the conservation of energy to replace $\omega$ we can rewrite this inequality as $\frac{u^2}{g R} \geq 2 - 3 \cos(\theta)$. For $\theta = \pi$ the right-hand side is maximal, so we obtain the second (stronger) condition $\frac{u^2}{g R} \geq 5 ~ \Leftrightarrow~ u \geq \sqrt{5gR}$.
For $2 \sqrt{gR} < u < \sqrt{5gR}$ the particle has enough energy to reach the top of the loop, but it will detach itself from the track before that happens. Since your entire calculation is based on the energy conservation law (which contains less information than the original equations of motion), the stronger condition for the applicability of your result never appears and must be found separately.
