Proof by induction that $\frac1{n+1}+ \frac1{n+2}+\cdots+\frac1{2n}=1-\frac{1}{2}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}.$ Prove that for any positive integer, $$\frac1{n+1}+ \frac1{n+2}+\cdots+\frac1{2n} = \left(1-\frac1{2}\right)+\left(\frac1{3}-\frac1{4}\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n}\right).$$
I have tried using a proof by induction but do not know how to approach the series. 
Note: This identity may be established by a clever algebraic manipulation, as seen here, but I am curious as to how an inductive proof might work.
 A: Hint:
To do the induction step first show that 
$$\frac{1}{2n+1}+\frac{1}{2n+2} - \frac{1}{n+1} = \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)$$
and then add this equation to 
$$\frac{1}{n+1} + \ldots + \frac{1}{2n} = \left(1-\frac{1}{2}\right) + \ldots + \left(\frac{1}{2n-1} - \frac{1}{2n}\right)$$
A: For each $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}.
$$
Note that the left-hand side constitutes the first $2n-1$ terms of what is called the alternating harmonic series. 
Base step ($n=1$): Notice that the left side of $S(n)$ has denominators which range from $1$ to $2n$, whereas the denominators on the right range from $n+1$ to $2n$. Thus, for $n=1$, the denominators on the left range from $1$ to $2$, whereas on the right, they range from $1+1=2$ to $2\cdot 1=2$; that is, there is only one term on the right. Consequently, $S(1)$ say that $1-\frac{1}{2}=\frac{1}{2}$, and this is true.
Inductive step: For some fixed $k\geq 1$, assume the inductive hypothesis $S(k)$
$$
S(k) : 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2k-1}-\frac{1}{2k}=\frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2k}
$$
to be true. We must then show that $S(k+1)$ follows:
$$
S(k+1) : \underbrace{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2k+1}-\frac{1}{2k+2}}_{\text{LHS}}=\underbrace{\frac{1}{k+2}+\frac{1}{k+3}+\cdots+\frac{1}{2k+2}}_{\text{RHS}}.
$$
Starting with the left-hand side of $S(k+1)$ (and filling two more penultimate terms),
\begin{align}
\text{LHS} &= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2}\\[1em]
  &= \frac{1}{k+1}+\frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2}\tag{by $S(k)$}\\[1em]
  &= \frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{k+1}-\frac{1}{2k+2}\\[1em]
  &= \frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{2}{2k+2}-\frac{1}{2k+2}\\[1em]
  &= \frac{1}{k+2}+\cdots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}\\[1em]
  &= \text{RHS},
\end{align}
we see that the right-hand side of $S(k+1)$ follows. This completes the inductive step $S(k)\to S(k+1)$. 
Hence, by mathematical induction, for all $n\geq 1, S(n)$ is true. $\blacksquare$
