TITLE: Prove $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3} +...+\frac{1}{2n}$
QUESTION [7 upvotes]: Prove that for all $n ≥ 1$
$$1 -
 \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}
 = \frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n}$$

My attempt:
By induction 
Base case: $n = 1$
LHS : $1 - \frac{1}{2} = \frac{1}{2}$
RHS : $\frac{1}{1 + 1 } = \frac{1}{2}$
Suppose equality holds for $n$
We need to show that it holds for $n + 1$, i.e 
$$1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}-+\cdots-\frac{1}{2n+2} = \frac{1}{n+2} + \frac{1}{n+3} +\frac{1}{n+4} +\cdots+\frac{1}{2n+2} $$
We have
$$\begin{align}
1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\cdots-\frac{1}{2n+2} & = \bigl[1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots - \frac{1}{2n}\bigr] + \frac{1}{2n+1} - \frac{1}{2n+2} \\
& = \frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\
&  =\frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} +\bigl (\frac{1}{n+1} - \frac{1}{2n+2} \bigr)\\
& = \frac{1}{n+2} +\frac{1}{n+3} +\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}
\end{align}$$ 
As desired. $\Box$
Is it correct?

REPLY [4 votes]: Your induction works.
As an alternative, 

writing $H_n =1 +\frac12 +\frac13 +\cdots +\frac1{n-1}+\frac1n$
then $H_{2n} =1 +\frac12 +\frac13 +\cdots +\frac1{2n-1}+\frac1{2n}$
while $\frac12H_n = \frac12 +\frac14 +\frac16 +\cdots +\frac1{2n-2}+\frac1{2n}$
so $H_{2n} - \frac12H_n =1 +\frac13 +\frac15 +\cdots +\frac1{2n-3} +\frac1{2n-1}$ 
and $H_{2n} - H_n =1 -\frac12 +\frac13 +\cdots +\frac1{2n-1}-\frac1{2n}$
which is equal to $H_{2n} - H_n = \frac{1}{n+1} +\frac{1}{n+2}+\frac{1}{n+3} +\cdots +\frac1{2n-1}+\frac1{2n}$

This then gives a reasonable approximation, since we know $H_n=\log_e{n}+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\cdots$ and so $H_{2n}=\log_e 2+\log_e{n}+\gamma+\frac{1}{4n}-\frac{1}{48n^2}+\frac{1}{1920n^4}-\cdots$ leading to
$H_{2n}-H_n = \log_e 2 - \frac{1}{4n}+\frac{1}{16n^2} -\frac{1}{128n^4} +\cdots$