Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem:

$$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$

 A: Hint:
$$\frac{1}{k(k+1)}= \frac{1}{k} - \frac{1}{k+1}.$$
Hint2:
$$\frac{n}{n+1} = 1 - \frac{1}{n+1}.$$
A: $\textbf{HINT:}$ Actually this is the answer in itself.
$\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$.
This is all that you will need when you do induction.
A: Every question of the form: prove by induction that
$$\sum_{k=1}^n f(k)=g(n)$$
can be done by verifying two facts about the functions
$f$ and $g$:


*

*$f(1)=g(1)$
and

*$g(n+1)-g(n)=f(n+1)$.

A: Essentially you test the base case, p(1), which is true since 1/(1+1) = 1/2.
Then, you want to assume your p(n) is true, namely, that sum you wrote above:
1/2 + .... + 1/(n+1)

and then add the next number in the sequence. 
1/[(n+1)+1]

Since you know your p(n) sum equals n/(n+1), you add the next number in the sequence above to your p(n):
1/2 + ... + 1/(n+1) + 1/[(n+1)+1] = p(n) + 1/[(n+1)+1] = n/(n+1) + 1/[(n+1)+1]
Given the useful hint above from user17762, they have done the algebra for you.
n/(n+1) + 1/[(n+1)+1] = (n+1)/(n+2)

You can rewrite as, (n+1)/(n+1+1), which is exactly p(n+1), so we assume it's true for all n because it's true for the next term in the sequence.
B.Y.U.
