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.

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 $n\geq 1$. $\blacksquare$

$$\begin{align} S&=\sum_{r=1}^n\frac 1{2r-1}-\sum_{r=1}^n\frac 1{2r}\color{orange}{+\sum_{r=1}^n\frac 1{2r}-\sum_{r=1}^n\frac 1{2r}}\\ &=\underbrace{\sum_{r=1}^n\frac 1{2r-1}+\sum_{r=1}^n\frac 1{2r}}_{}-\underbrace{2\sum_{r=1}^n\frac 1{2r}}\\ &=\qquad\quad\sum_{r=1}^{2n}\frac 1r\qquad \qquad-\sum_{r=1}^n\frac 1r\\ &=\color{blue}{\sum_{r=n+1}^{2n}\frac 1r}\qquad\blacksquare \end{align}$$

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, $$\require{cancel}\begin{align} S&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac1n\color{green}{+\frac 1{n+1}+\cdots-\frac 1{2n}}\\\\ &=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\ &\;- 2 \left(\frac12\qquad+\frac14\qquad+\frac16+\cdots+\frac 1n+\cdots\qquad\qquad+\frac 1{2n}\right)\\\\ &=\cancel{1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n}\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\ &\quad\;\;- \cancel{\left(1\qquad+\frac12\qquad+\frac13+\cdots+\frac 1{\frac12n}+\cdots\quad\qquad+\frac 1n\right)}\\\\ &=\color{blue}{\frac1{n+1}+\frac1{n+2}\cdots+\frac 1{2n}}\qquad\blacksquare \end{align}$$

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.

I would not separate the left side of the equation; I would work the proof from how the equation was given.

Proof: By mathematical induction.

P(1): $$1-\frac{1}{2} = \frac{1}{2} = \frac{1}{2} = \frac{1}{1+1}$$

P(n+1): Assume inductive hypothesis: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$$

We need to show the following:

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2(n+1)-1}-\frac{1}{2(n+1)} = \frac{1}{(n+1)+1}+\frac{1}{(n+1)+2}+...+\frac{1}{2(n+1)}$$

LHS: $$\begin{align}&1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2(n+1)-1}-\frac{1}{2(n+1)} =\\ & 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n+1}-\frac{1}{2n+2} =\\ & 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}+\frac{1}{2n+1}-\frac{1}{2n+2} =\\ & \left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}\right)+\frac{1}{2n+1}-\frac{1}{2n+2} =\\ & \left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)+\frac{1}{2n+1}-\frac{1}{2n+2} =\\ & \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}-\frac{1}{2n+2} =\\ & \frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{n+1}-\frac{1}{2n+2} =\\ & \frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{n+1}-\left(\frac{1}{2}\right)\frac{1}{n+1} =\\ & \frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\left(\frac{1}{2}\right)\frac{1}{n+1} =\\ & \frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2} =\\ & \frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+2} =\\ & \frac{1}{n+1+1}+\frac{1}{n+1+2}+...+\frac{1}{2(n+1)} =\\ & \frac{1}{(n+1)+1}+\frac{1}{(n+1)+2}+...+\frac{1}{2(n+1)} \\\end{align}$$

Therefore, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \forall n \in \mathbb N$. ∎