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

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


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.


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