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.

I'll give a hint that follows user43208's reasoning (though NB that the formula he gives is missing a numerical factor from one term and so leads to an incorrect result).
Hint The quantity $S_n$ can be written as
$$\left(\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + \cdots + \frac{1}{4 n - 1} + \frac{1}{4n}\right) \\ \quad- \left(\frac{1}{2n + 2} + \frac{1}{2 n + 4} + \cdots + \frac{1}{4 n - 2} + \frac{1}{4 n} \right),$$
and the second quantity in parentheses can be written as
$$\frac{1}{2}\left(\frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2 n -
 1} + \frac{1}{2 n}\right).$$

So we can write $$S_n = (H_{4n} - H_{2n}) - \tfrac{1}{2}(H_{2n} - H_n),$$ where $H_m$ is the $m$th harmonic number, $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{m}$. On the other hand, $H_n$ satisfies the asymptotic expansion $$H_m = \log m + \gamma + O\left(\frac{1}{m}\right) .$$ Here, $\gamma$ is the Euler-Mascheroni constant---in fact, it is usually essentially defined by this previous equation, but substituting the expansion in our above formula for $S_n$ shows that its value is irrelevant for purposes of evaluating our limit.

If the question had instead by tagged $\texttt{calculus}$, the intended method would presumably be to recognize $S_n$ as a Riemann sum, as Fedor Petrov's hint does. (NB the answer he states is also off by a numerical factor.)
