Recursion Relation - Solution

Question

Consider the relation $$p_x + p_{x-1}p_{x} = p_{x-1}$$ with $p_1$ given.

The solution is written $$\frac{1}{p_x} = \frac{1}{p_1} + (x-1)$$ in my lecture notes as though trivial.

I can't seem to get it.

I have tried to re-arrange the main expression for $\frac{1}{p_x} = 1 + \frac{1}{p_{x-1}}$, but I am not sure how to finish this off.

Any hint is appreciated

Answer 1

$\dfrac1{p_x}=\dfrac1{p_{x-1}}+1=\dfrac1{p_{x-2}}+2=\cdots=\dfrac1{p_1}+(x-1)$

Answer 2

Notice that sequence $a_x =\frac{1}{p_x}$ is arithmetic with difference $1$. So $$a_x = a_1+(x-1)\cdot 1$$ and thus a conclusion.

Answer 3

From

$$ p_k+p_kp_{k-1}-p_{k-1}=0\Rightarrow \frac{1}{p_{k-1}}+1-\frac{1}{p_k} = 0 $$

now calling $q_k = \frac{1}{p_k}$ we have

$$ q_{k-1} + 1 - q_{k} = 0 $$

with solution

$$ q_k = k+C_0 $$

hence

$$ p_k = \frac{1}{k+C_0} $$