Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I don't know how to find the sums of these sequences. I know the sums for odd and even integers, but I can't figure this out.
 A: Try using $\Sigma$-notation to make your problem more manageable in terms of its algebraic expressions and the like. To this end, note that 
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\tag{1}
$$
becomes
$$
\sum_{i=1}^n\frac{1}{2i-1}-\sum_{i=1}^n\frac{1}{2i}=\sum_{i=1}^n\frac{1}{i+n}.\tag{2}
$$
Before moving on to the induction proof, you should observe that
$$
\sum_{i=1}^{k+1}\frac{1}{i+k+1}=\sum_{i=2}^{k+2}\frac{1}{i+k},\tag{3}
$$
a simple $\Sigma$-manipulation that will be of use to us in a moment. Now let's prove $(2)$ (and hence $(1)$) by induction.

Claim: For $n\geq1$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^n\frac{1}{2i-1}-\sum_{i=1}^n\frac{1}{2i}=\sum_{i=1}^n\frac{1}{i+n}.
$$
Base step ($n=1$): $S(1)$ says that $\sum_{i=1}^1\frac{1}{2i-1}-\sum_{i=1}^1\frac{1}{2i}=\sum_{i=1}^1\frac{1}{i+1}$ and this is true because $1-\frac{1}{2}=\frac{1}{2}$. 
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 1$ and assume that 
$$
S(k) : \sum_{i=1}^k\frac{1}{2i-1}-\sum_{i=1}^k\frac{1}{2i}=\sum_{i=1}^k\frac{1}{i+k}
$$
holds. To be proved is that
$$
S(k+1) : \sum_{i=1}^{k+1}\frac{1}{2i-1}-\sum_{i=1}^{k+1}\frac{1}{2i}=\sum_{i=1}^{k+1}\frac{1}{i+k+1}
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{2i-1}-\sum_{i=1}^{k+1}\frac{1}{2i} &= \left(\sum_{i=1}^k\frac{1}{2i-1}-\sum_{i=1}^k\frac{1}{2i}\right)+\frac{1}{2k+1}-\frac{1}{2k+2}\tag{by defn.}\\[1em]
&= \sum_{i=1}^k\frac{1}{i+k}+\frac{1}{2k+1}-\frac{1}{2k+2}\tag{by $S(k)$}\\[1em]
&= \sum_{i=1}^{k+1}\frac{1}{i+k}-\frac{1}{2k+2}\tag{by defn.}\\[1em]
&= \sum_{i=2}^{k+2}\frac{1}{i+k}-\frac{1}{2k+2}-\frac{1}{2k+2}+\frac{1}{k+1}\tag{by defn.}\\[1em]
&= \sum_{i=2}^{k+2}\frac{1}{i+k}-\frac{1}{k+1}+\frac{1}{k+1}\tag{like terms}\\[1em]
&= \sum_{i=1}^{k+1}\frac{1}{i+k+1},\tag{by $(3)$}
\end{align}
one arrives at the right-hand side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step. 
Thus, by mathematical induction, the claim $S(n)$ is true for all $n\geq 1$. $\blacksquare$
A: $$\begin{align}
S&=\sum_{r=1}^n\frac 1{2r-1}-\sum_{r=1}^n\frac 1{2r}\color{orange}{+\sum_{r=1}^n\frac 1{2r}-\sum_{r=1}^n\frac 1{2r}}\\
&=\underbrace{\sum_{r=1}^n\frac 1{2r-1}+\sum_{r=1}^n\frac 1{2r}}_{}-\underbrace{2\sum_{r=1}^n\frac 1{2r}}\\
&=\qquad\quad\sum_{r=1}^{2n}\frac 1r\qquad \qquad-\sum_{r=1}^n\frac 1r\\
&=\color{blue}{\sum_{r=n+1}^{2n}\frac 1r}\qquad\blacksquare
\end{align}$$

The proof above is illustrated visually in the expansion below. 
If $n$ is odd, 
$$\require{cancel}\begin{align}
S&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots+\frac1n\color{green}{-\frac 1{n+1}+\cdots-\frac 1{2n}}\\\\
&=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\
&\;- 2 \left(\frac12\qquad+\frac14\qquad+\frac16+\cdots\qquad\;+\frac 1{n+1}+\;\cdots+\frac 1{2n}\right)\\\\\;
&=\cancel{1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n}\color{blue}{+\frac 1{n+1}+\cdots+\frac 1{2n}}\\
&\quad\;\;-  \cancel{\left(1\qquad+\frac12\qquad+\frac13+\cdots\qquad+\frac 1{\frac12(n+1)}+\cdots+\frac 1{n}\right)}\\\\
&=\color{blue}{\frac1{n+1}+\frac1{n+2}\cdots+\frac 1{2n}}\qquad\blacksquare
\end{align}$$
If $n$ is even, 
$$\require{cancel}\begin{align}
S&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac1n\color{green}{+\frac 1{n+1}+\cdots-\frac 1{2n}}\\\\
&=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\
&\;- 2 \left(\frac12\qquad+\frac14\qquad+\frac16+\cdots+\frac 1n+\cdots\qquad\qquad+\frac 1{2n}\right)\\\\
&=\cancel{1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n}\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\
&\quad\;\;-  \cancel{\left(1\qquad+\frac12\qquad+\frac13+\cdots+\frac 1{\frac12n}+\cdots\quad\qquad+\frac 1n\right)}\\\\
&=\color{blue}{\frac1{n+1}+\frac1{n+2}\cdots+\frac 1{2n}}\qquad\blacksquare
\end{align}$$
A: Denote $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$. What you want to prove is exactly
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} = H_{2n} - H_n
$$
i.e.,
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} - H_{2n} = - H_n
$$
Note that the left side is equal to
$$
2(-\frac{1}{2} - \cdots - \frac{1}{2n}) = -H_n
$$
since those terms in odd positions have been removed and those in even positions have been doubled.
