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 ?
 A: $$\sum_{i=n+2}^{2n+2} \frac{1}{i}=\sum_{i=n+1}^{2n}\frac{1}{i}+ \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$
It remains to prove that:
$$ \sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=1}^{2n+2}\frac{(-1)^{i+1}}{i}$$
that is
$$ \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{1}{2n+1}+\frac{-1}{2n+2} \Leftrightarrow \frac{2}{2n+2}=\frac{1}{n+1} (A).  $$
EDIT: If you want to continue in your way, you can do like this:
$$ \sum_{i=n+1}^{2n+1} \frac{1}{i} + \frac{-1}{2n+2}=\sum_{i=n+1}^{2n+1} \frac{1}{i}+ \frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=n+2}^{2n+2} \frac{1}{i}.$$
So,
$$ \sum_{i=n+2}^{2n+2} \frac{1}{i} =\sum_{i=1}^{2n+2} \frac{(-1)^{i+1}}{i}. $$
A: You got off on the wrong foot with your first step: you want to show that
$$\sum_{i=(n+1)+1}^{2(n+1)}\frac1i=\sum_{i=1}^{2(n+1)}\frac{(-1)^{1+i}}i\;,$$
i.e., that
$$\sum_{i=n+2}^{2n+2}\frac1i=\sum_{i=1}^{2n+2}\frac{(-1)^{1+i}}i\;.\tag{1}$$
In general I find it a little easier to start with this and figure out how to use the induction hypothesis than it is to start with the induction hypothesis and try to expand it somehow to get this. Also, I’d rather work with the lefthand side first, since it has slightly simpler terms. The lefthand side of the induction hypothesis is
$$\sum_{i=n+1}^{2n}\frac1i\;.$$
Compared with this, the lefthand side of the target equality, $(1)$, has two extra terms at the top, for $i=2n+1$ and $i=2n+2$, and it’s missing one at the bottom, for $i=n+1$. Thus,
$$\begin{align*}
\sum_{i=n+2}^{2n+2}\frac1i&=\sum_{i=n+1}^{2n}\frac1i+\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\\
&=\sum_{i=n+1}^{2n}\frac1i+\frac1{2n+1}+\frac1{n+1}\left(\frac12-1\right)\\
&=\sum_{i=n+1}^{2n}\frac1i+\frac1{2n+1}-\frac1{2n+2}\\
&=\sum_{i=1}^{2n}\frac{(-1)^{1+i}}i+\frac1{2n+1}-\frac1{2n+2}&\text{induction hyp.}\\
&=\sum_{i=1}^{2n}\frac{(-1)^{1+i}}i+\frac{(-1)^{1+(2n+1)}}{2n+1}+\frac{(-1)^{1+(2n+2)}}{2n+2}\\
&=\sum_{i=1}^{2n+2}\frac{(-1)^{1+i}}i\;,
\end{align*}$$
exactly as desired.
A: 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.
