We first show that

$$\lim_{n\to\infty}S_{2n}=\ln 2$$

Then because

$$\lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}\left(S_{2n}+\frac{1}{2n+1}\right)=\lim_{n\to\infty}S_{2n}=\ln 2$$

therefore we can conclude that 

$$\lim_{n\to\infty}S_{n}=\boxed{\ln 2}$$

To show $\displaystyle\lim_{n\to\infty}S_{2n}=\ln 2$, let's first transform $S_{2n}$:

$$\begin{align*} S_{2n} &= 1 - \frac{1}{2}+\frac{1}{3} - \frac{1}{4} +\cdots -\frac{1}{2n}\\ &= 1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} - 2\left(\frac{1}{2} + \frac{1}{4} + \cdots +\frac{1}{2n}\right)\\ &=1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} - \left(\frac{1}{1} + \frac{1}{2} + \cdots +\frac{1}{n}\right)\\ &= \frac{1}{n+1} +\frac{1}{n+2}+\cdots + \frac{1}{2n}\\ &= \frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots + \frac{1}{1+\frac{n}{n}}\right) \end{align*}$$

Applying the rectangular approximation model, the last expression is an approximation of $\frac{1}{1+x}$ between $(0, 1)$. Therefore,

$$\lim_{n\to\infty}S_{2n}=\int_0^1\frac{1}{1+x}dx =\int_0^1\frac{1}{1+x}d(x+1)=\ln (x+1)|_0^1 = \ln 2 $$