## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Since the given statement for $n=1$, and that when assumed to be true for a natural number k, the given statement is also true for $n=k+1$, then by the principle of Mathematical Induction, the given statement is true for all natural numbers $n$. (Refer to the step-by-step work below for the actual proof) Thus, for all natural numbers $n$, $\displaystyle \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+…+\frac{1}{n(n+1)}=\frac{n}{n+1}$
In order to prove the given statement, we will use the following two conditions for mathematical induction. a) The statement must hold for the first natural number, known as the base case. b) Assume that the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we will prove that the statement also holds for $n + 1$. This is known as the inductive step. So, we have: a) For $n=1$: $\dfrac{1}{2}=\dfrac{1}{1+1}=\dfrac{1}{2}$ b) Assume for $n=k$ : $\displaystyle \frac{1}{1(2)}+\frac{1}{2(3)}+…+\frac{1}{k(k+1)}=\frac{k}{k+1}$ Then for $n=k+1$: $\displaystyle\frac{1}{1(2)}+\frac{1}{2(3)}+…+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{(k+1)}+\frac{1}{(k+1)(k+2)}\\=\displaystyle\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\=\displaystyle\frac{k^2+2n+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}\\=\dfrac{(k+1)}{(k+2)}\\=\displaystyle\frac{(k+1)}{(k+1)+1}$ Thus we can see that the above two conditions are satisfied. So, by the Principal of Mathematical Induction, the given statement has been proved true for all natural numbers.