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.

$\displaystyle \sum\limits_{k=2n+1}^{4n-1} \frac{1}{k} - \sum\limits_{k=n+1}^{2n-1} \frac{1}{2k} \approx (\ln 4 – \ln 2) - \frac{1}{2}(\ln 2 – \ln 1) = \frac{\ln 2}{2} $ 

Explanation (with suitable variables $a,b,n,q,u,v$, e.g. $a\in\mathbb{N}$, $n\to\infty$):
$\displaystyle \prod\limits_{k=1}^{an} (1+\frac{x}{k}) \approx \frac{(an)^x}{\Gamma(1+x)}$ 
$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an}  (1+\frac{x}{k}) \approx (\frac{(an)^x}{\Gamma(1+x)})/(\frac{n^x}{\Gamma(1+x)}) = a^x $
$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an}  (1+\frac{x}{qk})^q \approx a^x $ , $\hspace{2cm} q\to\infty$ :
$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an}  \exp(\frac{x}{k}) \approx a^x$ , $\hspace{2cm} log$ :
$=> \hspace{1cm}\displaystyle \sum\limits_{k=n}^{an}  \frac{1}{k} \approx \ln a  $
$=> \hspace{1cm}\displaystyle \sum\limits_{k=bn+u}^{an+v} \frac{1}{k} \approx \ln a - \ln b \enspace\enspace$   with  constants $\enspace|u|,|v|\ll n $    
