Find $\lim{n \to \infty}\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}$

Question

I've proved that $\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}$, but I have no idea how to sum up this formula and take limit. Can anyone give me some tips? Thanks a lot!

UPD: Our instructor says that we can't use tools that are not taught in class before and I came up with another idea.

It seems that $\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}$,

and $\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}=(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n})$ - $(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n})$ and it is about $\ln(2n) - \ln(n)$ and thus the answer is $\ln(2)$. Am I consider it right?

Answer 1

HINT Consider the Taylor series for the natural logarithm function around $x=1$.

Answer 2

As mentioned in the comments, I would first rewrite the sum as

$$\sum_{m=1}^{2n} \frac{(-1)^{m-1}}{m}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots+\frac{1}{2n-1}-\frac{1}{2n}\tag{1}$$

then recall the Taylor series for $\ln(x+1)$ $$\ln(x+1)=\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m}x^m=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\dots\tag{2}$$

If you let $n\to\infty$ in $(1)$, then you can match the alternating series in $(1)$ and $(2)$ by choosing a suitable value for $x$.