If I solve it by a Riemann sum approach i.e. $$\int_{1}^{\infty} 1/x^2 dx$$ then I can see that the sum of the geometric series is equals to 1. He mentioned that it should not be equals to 1, but instead the inequality $\leq 1$. It made sense to me at first, but after that it seemed to be even more illogical.
The fact that the first term itself is 1, and you are adding on 1/4, 1/9 .... should make it strictly more than 1. So how can it be $\leq 1$?
The integral $\int_{1}^{\infty} 1/x^2 dx$ does not equal the sum $\sum_{n = 1}^{\infty} 1/n^2$.
Evaluating the integral gives a convergence to the value 1. Think of this integral as taking the area under the curve $1/x^2$ from x=1 to x=$\infty$. This convergence can be observed through the p-test, where p is the power that x is raised to. If p > 1, then the series converges. The exact value of convergence can be calculated by taking the antiderative and plugging in the limits. $$\int_{1}^{\infty} 1/x^2 dx = -\frac{1}{x}\Big|_1^\infty = -\frac{1}{\infty}-(-1) = 0 +1 = 1$$
The sum $\sum_{n = 1}^{\infty} 1/n^2$ evaluates to $\frac{{\pi}^2}{6} \approx 1.644934$. This is what you are thinking of when you mention taking 1 as the first term, then adding 1/4, 1/9... This is the famous Basel problem that Euler solved in the 1700's. A simple proof (Euler's approach) can be found here.