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.

The sum can be written as
$$S(n) = \sum_{k=n}^{2n} \frac{1}{2k-1}$$
If we replace the sum by an integral we find
$$I(n) = \int_n^{2 n} \frac{1}{2 k-1} \, dk = \frac{1}{2} \log (2+\frac{1}{2n-1})$$
And the limit $n\to\infty$ is
$$S(n\to\infty) = \frac{1}{2} \log (2)\tag{1}$$
The justification of the replacement of the sum by an integral is easily done considering the inequalities (lower "sum" < S < upper "sum")
$$\int_n^{2 n} \frac{1}{2 k-1} \, dk < S(n) < \int_n^{2 n} \frac{1}{2 (k-1)-1} \, dk $$
which are obvious graphically.
Hence we have
$$ \frac{1}{2} \log \left(2+\frac{1}{2n-1}\right)< S(n) < \frac{1}{2} \log \left(2 + \frac{3}{2 n-3}\right)$$
and the limit $n\to\infty$ gives back the result (1).
Generalization
It is tempting to consider the more general sum (where $0<a<b, c<0, d<a \;c \;n$)
$$S(a,b,c,d,n)=\sum _{k=a\; n}^{b\; n} \frac{1}{c\; k-d}$$
The limit is similarly calculated with the result
$$\lim_{n\to \infty } \, S(a,b,c,d,n)=\frac{1}{c}\;\log \left(\frac{b}{a}\right)$$
Notice the independence of $d$.
