1.17

Step 1: Show that the statement holds for the basis case $n=1$

${\frac {1}{i(i+1)}}={\frac {n}{n+1}}$

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

${\frac {1}{2}}={\frac {1}{2}}$

Since $1/2=1/2$, the basis case is true.

Step 2: Assume that that summation is true up to n.

Step 3: Show that on the assumption that the summation is true for n, it follows that it is true for n + 1.

$\sum _{i=1}^{n+1}={\frac {n+1}{n+1+1}}={\frac {n}{n+1}}+{\frac {1}{(n+1)(n+1+1)}}$
${\frac {n+1}{n+2}}={\frac {n(n+2)}{(n+1)(n+2)}}+{\frac {1}{(n+1)(n+2)}}$
${\frac {n+1}{n+2}}={\frac {n(n+2)+1}{(n+1)(n+2)}}$
${\frac {n+1}{n+2}}={\frac {n^{2}+2n+1}{(n+1)(n+2)}}$
${\frac {n+1}{n+2}}={\frac {(n+1)(n+1)}{(n+1)(n+2)}}$
${\frac {n+1}{n+2}}={\frac {(n+1)}{(n+2)}}$

QED

Back to Chapter 1.