Consider a series, $$S=\sum_{n\ge0}\frac{(-1)^n}{2n+1}$$
Which can also be written as, $$S=\lim_{n\rightarrow\infty}\sum_{r=0}^{n} \frac{2n}{(4r+1)(4r+3)}\cdot \frac{1}{n}$$ Substituting $$\frac{r}{n}=x,r=nx,\frac{1}{n}=dx$$ Gives,
$$S=\int_{0}^{1}\frac{2n}{(4nx+1)(4nx+3)}\cdot dx$$
Which gives, $$S=\frac{\log(12n+3)-\log(4n+3)}{4}$$
As $n$ goes to infinity $S$ goes to $\frac{\log{3}}{4}$.
But this page states the sum to be $\frac{\pi}{4}$. I can't figure it out that what am I missing in this approach? Can it be solved using Riemann's sum?
There is no reason to think that a sequence of Riemann sums for different functions converge, even if $\Delta x\to 0.$
Here, you have a sequence of functions $f_n(x)=\frac{2n}{(4nx+1)(4nx+3)}.$ And you have the Riemann sum of $f_n$ for $\Delta x=\frac1n.$
If we write $S_{n,m}$ as the Riemann sum for $f_n$ with $\Delta x=\frac1m,$ then you want:
$$\lim_{n\to\infty} S_{n,n}.\tag1$$
Now, what we know is that, for each $n,$ $$\lim_{m\to\infty} S_{n,m}=\int_0^1 f_n(x)\,dx\tag2$$
But there is no reason to think that helps us compute the limit $(1).$ This is because the convergences in $(2)$ are not uniform. In particular, since $f_{n}(0)=\frac{2n}3\to\infty,$ we need $\Delta x$ to be small relative to $f_n(0)$ to get close to the limit integral.
What you are essentially saying is that if we have a double sequence $a_{n,m}$ and a sequence $b_n=\lim_{m\to\infty} a_{n,m},$ then $$\lim_{n\to\infty} a_{n,n} =\lim_{n\to\infty} b_n.$$
But this isn't true in general, as can be seen with $a_{n,m}=2^{n-m}.$
Then $\lim_m a_{n,m}=0$ for each $n,$ but $\lim_n a_{n,n}=1.$
You can only do this if the limits converging to $b_n$ are uniform. But there is no reason to think your integrals converge uniformly, and, due to the unbounded nature of $f_n(0),$ actual reason to believe they are not.