Let the property \(P(k)\) be \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}=\frac{k}{k+1}\)

Show that \(P(k)\) is true for all integers \(k\geq1\) using mathematical induction

Basis Step: \(P(k)\) is true:

That is to show that \(\frac{1}{1\cdot2}=\frac{1}{1+1}\)

The left hand side of the equation is \(\frac{1}{1\cdot2}=\frac{1}{2}\) and right-hand side is

\(\frac{1}{1+1}=\frac{1}{2}\)

It follows that \(\frac{1}{2}=\frac{1}{2}\)

Hence \(P(1)\) is true.

Show that for all integers \(n\geq1\), \(P(k)\) is true then \(P(k+1)\) is also true:

Suppose P(k) is true.

Then the inductive hypothesis is

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}\)

Now show that \(P(k+1)\) is true.

That is to show that

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}\)

Or, equivalently that

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}\)

The left-hand side of \(P(k+1)\) is

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}\)

\(=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\)

\(=\frac{1}{k+1}(k+\frac{1}{k+2})\)

\(=\frac{1}{k+1}(\frac{k^2+2k+1}{k+2})\)

\(=\frac{1}{k+1}(\frac{(k+1)^2}{k+2})\)

\(=\frac{1}{k+1}\frac{(k+1)^2}{k+2}\)

\(=\frac{k+1}{k+2}\)

which is right hand side of \(P(k+1)\)

Hence from the principle of mathematical induction,

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}\) is true, for all integers \(n\geq1\)