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$$\lim_{n\to\infty}\frac {1-\frac {1}{2} + \frac {1}{3} -\frac {1}{4}+ ... + \frac {1}{2n-1}-\frac{1}{2n}}{\frac {1}{n+1} + \frac {1}{n+2} + \frac {1}{n+3} + ... + \frac {1}{2n}}$$
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.
Here's a solution with no calculus, just algebra. The numerator and the denominator are actually equal for all $n.$
Let $H_n=\sum_{k=1}^n \frac1{k}$ be the $k^{\text{th}}$ harmonic number.
Your denominator is $$D_n=\big(1+\frac12+\frac13+\dots+\frac1{2n}\big)-\big(1+\frac12+\frac13+\dots+\frac1{n}\big)=H_{2n}-H_n.$$
Your numerator is $$N_n=1-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}=\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}.$$
Compute
\begin{align}H_{2n}-N_n &=\big(1+\frac12+\frac13+\dots+\frac1{2n}\big)-\big(1-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}\big) \\ &= 2\cdot\!\!\!\!\!\sum_{\substack{1\le k \le 2n\\k\text{ is even}}}\frac1{k} \scriptsize\quad\quad{\text{ (because the odd terms cancel out and the even terms are doubled up)}} \\&=2\cdot\sum_{j=1}^n\frac1{2j} \\&=\sum_{j=1}^n \frac1{j} \\&=H_{n}, \end{align}
so $$ N_n=H_{2n}-H_n.$$
It follows that $N_n=D_n$ for all $n,$ so $N_n/D_n$ is a constant sequence with value always $1,$ and the limit is therefore $1.$
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:
$$\frac{\ln 2}{\ln 2}=1$$
