Hello WartonMorton Originally Posted by

**WartonMorton** Please see attachment.

I know that math induction is when you prove the first term and then you have to prove for all the other terms with some sort of k+1? It's in a little bit different format than I'm used to so maybe that's why I am getting confused.

The problem is this:

Use mathematical induction to prove the following assertion:

If $\displaystyle a_1 = 1$ and $\displaystyle a_{n+1}=a_n - \frac{1}{n(n+1)}$, then $\displaystyle a_n = \frac1n$.

The thing you have to prove here is that $\displaystyle a_n = \frac1n$. This forms our 'inductive assertion' or 'induction hypothesis', that we can denote by $\displaystyle P(n)$.

So $\displaystyle P(n)$ is the statement: $\displaystyle a_n = \frac1n$.

We have no means of telling whether $\displaystyle P(n)$ is true yet. It is just a hypothesis. What we can do, though, is to make use of the definition of $\displaystyle a_{n+1}$ to deduce another statement that will follow if $\displaystyle P(n)$ is true. From this we hope to prove that $\displaystyle P(n) \Rightarrow P(n+1)$ (where $\displaystyle \Rightarrow$ means 'implies'), which is the basis of a proof by induction.

So we use the 'implies' sign, when we make this deduction. Like this:$\displaystyle P(n) \Rightarrow a_n = \frac1n$$\displaystyle \Rightarrow a_{n+1} = \frac1n-\frac{1}{n(n+1)}$, using the definition of $\displaystyle a_n+1$ that we've been given.$\displaystyle =\frac{(n+1) - 1}{n(n+1)}$

$\displaystyle =\frac{1}{n+1}$

$\displaystyle \Rightarrow P(n+1)$

Do you understand that last step? We have shown that if $\displaystyle P(n)$ is true, then $\displaystyle a_{n+1} = \frac{1}{n+1}$, which is simply $\displaystyle P(n+1)$.

Finally we check that $\displaystyle P(1)$ is true. Which it is:

$\displaystyle a_1 = 1 = \frac{1}{1}$

This completes the proof.

Grandad