Proving the summation formula using induction: $\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$ I am trying to prove the summation formula using induction:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$
So far I have...
Base case:


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*Let n=1 and test
$\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
$\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$  
$\frac{1}{2} = \frac{1}{2}$


*

*True for n=1
Induction Hypothesis:


*

*Assume the statement is true for the n-th case
$\sum_{k=1}^n  \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
Inductive Step:


*

*Prove, using the Inductive Hypothesis as a premise, that
$$\sum_{k=1}^{n+1}  \frac{1}{k(k+1)} = \sum_{k=1}^{n}  \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$
To prove
$$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$
Multiply both sides by $n+2$ to get an equivalent expression.
$$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2)  $$
$$ n+1=n+2−1 $$
Does this all make sense? How can this be improved upon?

$$\sum_{k=1}^{m+1} \frac1{k(k+1)} = \sum_{k=1}^m\frac1{k(k+1)}  +\frac1{(m+1)(m+2)}$$
$$ = 1-\frac1{m+1}+\frac1{(m+1)(m+2)}=1-\frac{m+2-1}{(m+2)(m+1)}=\cdots$$
