The task at hand is to prove using induction that the following proposition holds for all $n \in \mathbb{N}$.
$$P(n): \sum_{k=1}^{n}1/k(k+1) = n/(n+1)$$
Here is the proof I have thus far:
Base Case: $(n=1)$
LHS $= 1/1(1+1) = 1/(1+1) =$ RHS
i.e., both LHS and RHS are $1/2$.
So the base case is true.
Step Case: Assume the proposition holds $n=k$; then show that it holds for $n=k+1$.
$1/k(k+1) + 1/(k+1)(k+2)$
$= k/(k+1) + 1/(k+1)(k+2)$
$= k(k+2)/(k+1)(k+2) + 1/(k+1)(k+2)$
$= k(k+2)+ 1/(k+1)(k+2)$
$= k^2 + 2k + 1$
$= (k + 1)^2 / (k+1) (k+2)$
$= (k+1)/(k+2)$
$= (k+1)/((k+1)+1)$
Is this the right way to do it?
If so, where could my proof be clearer?
If not, could you help me to find the error(s)?
Best Answer
I think you mean the $\sum{\frac{1}{k(k+1)}}$
If so: $$\sum_{1}^{n}{\frac{1}{k(k+1)}} = \sum_{1}^{n}{\frac{1}{k} -\frac{1}{k+1}} = \frac{n}{n+1}$$
But if you want induction:
Step: $$\sum_{1}^{n+1}\frac{1}{k(k+1)} = \sum_{1}^{n}{\frac{1}{k(k+1)}} + \frac{1}{(n+1)(n+2)} = \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$$