# Prove explicit form of a recursive sequence [closed]

Let the sequence $A_{n+1}$: $$A_{n+1}=\frac{A_n+\lambda}{1+A_n}, n=0,1,2\dots, \lambda>0$$

Supose $A_0=a_0>0.$

Let the sequence $B_{n}$:

$$B_{n}=\frac{1}{A_n+\sqrt \lambda}$$

Check $\{B_n\}^\infty _{n=0}$ is the explicit form of the following recursive sequence:

$$\left \{ B_0 = b_0 \atop B_{n+1} = rB_n+b, n=0,1,2\dots \right.$$ for some $b_0,r,b$ that depends of $a_0$ and $\lambda$

I am not sure how solve this problem, I tried to use induction but it doesn't works well, so can you help me with this please?

## closed as off-topic by John B, Namaste, Chris Custer, JonMark Perry, GNUSupporter 8964民主女神 地下教會Apr 24 '18 at 14:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, Namaste, Chris Custer, JonMark Perry, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.

## 3 Answers

$$B_{n+1} = \frac 1{A_{n+1} + \sqrt \lambda } = \frac 1{\frac{A_n+\lambda}{1+A_n} + \sqrt \lambda} = \frac{1+A_n}{ A_n + \lambda + (1+A_n)\sqrt \lambda} = \frac{1+A_n}{A_n(1+\sqrt \lambda) + \sqrt \lambda (1+\sqrt \lambda)} = \frac{1+A_n}{(A_n+\sqrt \lambda)(1+\sqrt\lambda)} = B_n\frac{\frac {1}{B_n} - \sqrt\lambda+1}{1+\sqrt\lambda} = \frac{1-\sqrt \lambda}{1+\sqrt\lambda} B_n + \frac1{1+\sqrt\lambda}$$

Hint: $\displaystyle\; B_{n}=\frac{1}{A_n+\sqrt \lambda} \iff A_n=\frac{1}{B_n} - \sqrt{\lambda}\,$, then replace in the original recurrence:

$$A_{n+1}=\frac{A_n+\lambda}{1+A_n} \quad\iff\quad \cfrac{1}{B_{n+1}} - \sqrt{\lambda} = \frac{\cfrac{1}{B_n} - \sqrt{\lambda}+\lambda}{1+\cfrac{1}{B_n} - \sqrt{\lambda}}$$

Isolate $B_{n+1}$ on one side, simplify the expression etc, and you get the recurrence for $\,B_n\,$.

You don't need induction.

$$B_0 = b_0$$

Trivially true, since it only has to be true for some $b_0$, and everything is equal to something.

$$B_{n+1} = rB_n + b$$

They key here is you want to find an $r$ and $b$ that does not depend on $n$. The logic here is utilizing: "if the subsequent equation has a $r,b$ that works, then the prior one does as well". Plug in the definition of $B$:

$$\frac{1}{A_{n + 1} + \sqrt \lambda} = r \frac{1}{A_{n} + \sqrt \lambda} + b$$

Plug in the recursive relation to see if you get something of the right form:

$$\dfrac{1}{\left(\dfrac{\lambda + A_n}{1 + A_n}\right) + \sqrt \lambda} = r \frac{1}{A_n + \sqrt \lambda} + b$$

All the rest is just simplifying fractions.

$$\dfrac{1}{\left(\dfrac{\lambda + A_n}{1 + A_n}\right) + \left(\dfrac{1 + A_n}{1 + A_n}\right)\sqrt \lambda} = \frac{r}{A_n + \sqrt \lambda} + \frac{A_n + \sqrt \lambda}{A_n + \sqrt \lambda}b$$

$$\dfrac{1}{\left(\dfrac{\lambda + A_n + \sqrt \lambda + \sqrt \lambda A_n}{1 + A_n}\right)} = \frac{r + bA_n + \sqrt \lambda}{A_n + \sqrt \lambda}$$

$$\dfrac{1 + A_n}{A_n(1 + \sqrt \lambda) + \lambda + \sqrt \lambda} = \frac{r + \sqrt \lambda + bA_n}{A_n + \sqrt \lambda}$$

Then make the denominators match:

$$\dfrac{1 + A_n}{A_n(1 + \sqrt \lambda) + \lambda + \sqrt \lambda} = \left(\frac{1 + \sqrt \lambda}{1 + \sqrt \lambda}\right)\left(\frac{r + \sqrt \lambda + bA_n}{A_n + \sqrt \lambda}\right)$$

$$\dfrac{1 + A_n}{A_n(1 + \sqrt \lambda) + \lambda + \sqrt \lambda} = \frac{(1 + \sqrt \lambda)(r + \sqrt \lambda) + b(1 + \sqrt \lambda)A_n}{A_n(1 + \sqrt \lambda) + \lambda + \sqrt \lambda}$$

So we can see now that there is a choice of $r$ and $b$ that makes the numerators equal also. Since every step along the way was an equivalence, the presence of those $r,b$ here makes it also hold in the original equation.