Having difficulty solving a recurrence relation

Question

Given that $a_{n+1}=\dfrac{a_{n-1}}{1+n\cdot a_{n-1}a_{n}}$ for $n=1,2,3,....$ and $a_0=a_1=1$. Find the value of $a_{199}\cdot a_{200}$. Also give with proof the general formula of $a_{n}a_{n+1}$?

Answer 1

It is easy enough to confirm your formula.

Define $$b_n=\frac 1{a_na_{n-1}}-1$$

We have $b_1=0$ and we wish to show that $$b_n=b_{n-1}+n-1$$

You have checked that equality holds for small $n$, let's proceed by induction.

But we write out both sides to see that we are trying to show $$\frac 1{a_na_{n-1}}=\frac 1{a_{n-1}a_{n-2}}+n-1$$

Clearing denominators gives us $$a_{n-2}=a_n+a_na_{n-1}a_{n-2}(n-1)=a_n(1+a_{n-1}a_{n-2}(n-1))$$ $$\iff a_n=\frac {a_{n-2}}{1+a_{n-1}a_{n-2}(n-1)}$$

which is the defining relation for the $a_i$. And we are done.

Answer 2

I did the find the general formula for $a_{n}\cdot a_{n+1}$ but I do not have a proof.

1. $\dfrac{1}{a_0a_1}=1$
2. $\dfrac{1}{a_1a_{2}}=2$
3. $\dfrac{1}{a_2a_3}=4$
4. $\dfrac{1}{a_3a_4}=7$
5. $\dfrac{1}{a_4a_5}=11$

I think it is trivial to see the pattern.

$$\dfrac{1}{a_{n}\cdot a_{n+1}}=1+(1+2+3....n)=1+\dfrac{n(n+1)}{2}$$

But however as I earlier said, I don't have a sound proof.

Answer 3

You can prove the formula you found, by induction:

given $a_{n+1}=\dfrac{a_{n-1}}{1+n\cdot a_{n-1}a_{n}}$, it follows that $\dfrac1{a_{n+1}a_n}=\dfrac1{a_{n-1}a_n}+n.$