6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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. $\blacksquare$

$\endgroup$
3
$\begingroup$

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

$$\frac{1}{n+1} + \ldots + \frac{1}{2n} = \left(1-\frac{1}{2}\right) + \ldots + \left(\frac{1}{2n-1} - \frac{1}{2n}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.