Question

If $\underset{n\to \infty }{lim}(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+\cdots +\frac{1}{\sqrt{3}n})=\frac{\pi }{a},$ then the unit digit of $(a^{2020}+2019)$ is

Answer 1

$L=\underset{n\to \infty }{lim}(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+\cdots +\frac{1}{\sqrt{3}n})$
The given limit can be writen as
$\underset{n\to \infty }{lim}(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+\cdots +\frac{1}{\sqrt{4n^2-n^2}})\Rightarrow L=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{\sqrt{4n^2-r^2}}=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n\sqrt{4-\frac{r^2}{n^2}}}$
From the theory of integration of limit as a sum,
$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}f(\frac{r}{n})=\underset{0}{\overset{1}{\int }}f\left(x\right)dx\Rightarrow L=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{4-x^2}}={\left[{\sin }^{-1}(\frac{x}{2})\right]}_0^1\Rightarrow L=\frac{\pi }{6}=\frac{\pi }{a}\Rightarrow a=6$

We know that the unit digit of $6^{2020}$ is $6.$
So, the unit digit of $6^{2020}+2019$ will be $5$.