Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere between 0.5 and 1, but that's not the exact answer. 
Manually doing fist elements gives 
$$\frac{1}{3}, \quad \frac{1}{5} + \frac{1}{7}, \quad \frac{1}{7} + \frac{1}{9} + \frac{1}{11}, \quad ...$$
I'm trying to restate it as 
$$\lim_{n \to \infty} \bigg[ \sum_{i = 1}^{\infty} \frac{1}{2i - 1} - \sum_{i = 1}^{n} \frac{1}{2i - 1} - \sum_{i = 4n}^{\infty} \frac{1}{2i - 1} \bigg]$$
so that I cut off the beginning of the series and it's tail.

Consider the following inequalities:
$$ \begin{align}
 \frac{1}{2n+1+2x} &\le \frac{1}{2n+1} \le \frac{1}{2n-1+2x}, \quad 0\le x\le 1, \quad (1) \\
 \frac{1}{2n+1+2x} &\le \frac{1}{2n+3} \le \frac{1}{2n-1+2x}, \quad 1\le x\le 2 , \quad (2)\\
 \frac{1}{2n+1+2x} &\le \frac{1}{2n+5} \le \frac{1}{2n-1+2x}, \quad 2\le x\le 3 , \quad (3)\\
 \frac{1}{2n+1+2x} &\le \frac{1}{2n+7} \le \frac{1}{2n-1+2x}, \quad 3\le x\le 4 , \quad (4)\\
... \quad & \le \quad ... \quad \le \quad... \\
 \frac{1}{2n+1+2x} &\le \frac{1}{2n+2n-1} \le \frac{1}{2n-1+2x}, \quad n-1\le x\le n. \quad (5)
\end{align} $$
Integrating each inequality over the corresponding interval, side by side.
For example:
$$ \begin{align}
 \int_0^1 (1) dx \Rightarrow \int_0^1 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+1} \le \int_0^1 \frac{dx}{2n-1+2x}, \quad (1') \\
 \int_1^2 (2) dx \Rightarrow \int_1^2 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+3} \le \int_1^2 \frac{dx}{2n-1+2x}, \quad (2') \\
 \int_2^3 (3) dx \Rightarrow \int_2^3 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+5} \le \int_2^3 \frac{dx}{2n-1+2x}, \quad (3') \\
... \quad & \le \quad ... \quad \le \quad... \\
\int_{n-1}^n (5) dx \Rightarrow \int_{n-1}^n \frac{dx}{2n+1+2x} &\le \frac{1}{2n+2n-1} \le \int_{n-1}^n \frac{dx}{2n-1+2x}. \quad (5') 
\end{align} $$
Summing $(1')$ through $(5')$, side by side:
$$ \begin{align}\int_0^n \frac{dx}{2n+1+2x} &\le S_n \le \int_0^n \frac{dx}{2n-1+2x} \quad (6)\\
\frac12 \ln|2n+1+2x|_0^n &\le S_n \le \frac12 \ln |2n-1+2x|_0^n \\
\frac12 \ln\left|\frac{4n+1}{2n+1}\right| &\le S_n \le \frac12 \ln \left|\frac{4n-1}{2n-1}\right|, \quad (7)
\end{align}$$
Thus, from $(7)$:
$$ \lim_{n\to\infty} S_n = \frac{\ln 2}{2} $$
