<s>[INST] 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.
 [/INST] 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:
$$\frac{\ln 2}{\ln 2}=1$$
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