Having problem in last step on proving by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$ The question I am asked is to prove by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$ 
its easy to prove this holds for $n =1$ that gives $\frac{1}{2}=\frac{1}{2}$
Now assuming $n$ its true I want to say it is true $n+1$. So,
$\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ 
$\sum^{2n}_{i=n+1}\frac{1}{i}+\frac{(-1)^{1+(2n+1)}}{2n+1}+\frac{(-1)^{1+(2n+2)}}{2n+2}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i}+\frac{(-1)^{1+(2n+1)}}{2n+1}+\frac{(-1)^{1+(2n+2)}}{2n+2} $ 
$\sum^{2n}_{i=n+1}\frac{1}{i}+(-1)^{2n+2}[ \frac{1}{2n+1}+\frac{(-1)}{2n+2}]=\sum^{2n+2}_{i=1}\frac{(-1)^{1+i}}{i} $ 
$\sum^{2n}_{i=n+1}\frac{1}{i}+ \frac{1}{2n+1}+\frac{(-1)}{2n+2}=\sum^{2(n+1)}_{i=1}\frac{(-1)^{1+i}}{i} $ 
$\sum^{2n+1}_{i=n+1}\frac{1}{i}+\frac{(-1)}{2n+2}=\sum^{2(n+1)}_{i=1}\frac{(-1)^{1+i}}{i} $ 
i don't know what can i do next if the numerator of  $\frac{1}{i}+\frac{(-1)}{2n+2}$ was positive i knew. is there a way i can turn it positive?or my approach is wrong ?

Since an answer how to continue your proof was already given, I want to give another proof not using induction (which is in my eyes not the right way to proof this equality).
We have
$$\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i} \overset{(1)}{=} \sum_{i=1}^{n}\frac{1}{2i-1}-\sum_{i=1}^{n}\frac{1}{2i} = \sum_{i=1}^{n}\frac{1}{2i-1}+\sum_{i=1}^{n}\frac{1}{2i}-2\sum_{i=1}^{n}\frac{1}{2i}$$
$$\overset{(2)}{=}\sum_{i=1}^{2n}\frac{1}{i}-\sum_{i=1}^{n}\frac{1}{i} = \sum_{i=n+1}^{2n}\frac{1}{i},$$
where $(1)$ is dividing the sum into even and odd part and (2) is the inverse.
