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 $u\geq \sqrt{5gR}$

Call problem $A$ the one you have drawn: the tracks may only provide a normal force radially inwards. Call problem $B$ the similar problem except that cart and track are replaced by bead on a wire. In $A$, the cart may fall off the track. In $B$, the 'cart' may not fall off the 'track'.
For problem $A$:
Minimum initial speed to reach height of $2R$: at least $\sqrt{4gR}$ (exact number not relevant to the arguments here)
Minimum initial speed to complete a loop: $\sqrt{5gR}$.
For problem $B$:
Minimum initial speed to reach height of $2R$: $\sqrt{4gR}$.
Minimum initial speed to complete a loop: $\sqrt{4gR}$.
Your equations of motion are those of problem $B$, since you have fixed $r=R$. The problems are identical when $u>\sqrt{5gR}$, when the normal force of the track is inwards over the whole loop. The problems are not identical if $u<\sqrt{5gR}$.
You get sensible answers (real and finite integral) for $\sqrt{4gR}<u<\sqrt{5gR}$, because in $B$, the cart does complete the loop. We know that in $A$, the cart does not for $u<\sqrt{5gR}$, so we have found sensible answers to a different problem.
Your elliptic integral has a singularity at $\sqrt{4gR}$. Wolfram agrees. This indicates $T \to \infty$ in problem $B$, as $u \to \sqrt{4gR}$: the bead takes infinitely long to reach the top of the loop.
