Find the limit of the following expression: $$\lim_{n\to\infty}\frac {1-\frac {1}{2} + \frac {1}{3} -\frac {1}{4}+ ... + \frac {1}{2n-1}-\frac{1}{2n}}{\frac {1}{n+1} + \frac {1}{n+2} + \frac {1}{n+3} + ... + \frac {1}{2n}}$$
I can express the value of the geometric sum of ${\frac {1}{2} + \frac {1}{4}+...+\frac {1}{2n}}$ but the others are ahead of me. 
Putting both fraction parts under a common denominator makes that part tidy but the numerator seems to get way too complicated, which makes me think there is some simple way to do this.

In the denominator we have:
$$\sum_{i=1}^{n} \frac{1}{n+i}$$
And as $n \to \infty$ this is:
$$\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n+i}=\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{1+(0+i \frac{1-0}{n})}\frac{1-0}{n}=\int_{0}^{1} \frac{1}{1+x} dx=\ln (1+1)-\ln(1+0)=\ln (2)$$
For the numerator as $n \to \infty$, considering the geometric series $\sum_{n=0}^{\infty} u^n=\frac{1}{1-u}$ for $|u|<1$, substituting in $-x$ for $u$ we have:
$$\frac{1}{1-(-x)}=\sum_{n=0}^{\infty} (-1)^nx^n$$
Then integrate both sides to get
$$\ln(1+x)=C+\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n+1}}{n+1}=C+\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
Let $x=0 \implies C=0$, so:
$$\ln (1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
Because the series converges for $x=1$ by the alternating series test, we are justified in letting $x \to 1^-$ to get:
$$\ln (2)=\sum_{n=1}^{\infty} (-1)^n\frac{1}{n}$$
Hence the limit is:
$$\frac{\ln 2}{\ln 2}=1$$
