Show the convergence of the recursive sequence without computing it's limit

Question

$a_0=2, \quad a_{n + 1} = \frac{2a_n}{1+a_n}$

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So I have started to compute the first elements:

$\quad a_{0} = 2$

$\quad a_{1} = \frac{4}{3}$

$\quad a_{2} = \frac{8}{7}$

$\quad a_{3} = \frac{16}{15}$

$\quad a_{4} = \frac{32}{31}$

So the first elements show that $\quad a_{n} = \frac{2^{n+1}}{2^{n+1}-1}$

My conjecture is that $\quad a_{n} = \frac{2^{n+1}}{2^{n+1}-1}$ true is and I can prove it by using induction.

for $\quad a_{0} = 2$ it works

$\quad a_{n} = \frac{2^{n+1}}{2^{n+1}-1}=\quad a_{n} = \frac{2\cdot2^{n}}{2\cdot2^{n}-1}$

and I'm struggling at this point how can I continue it? Thanks in advance a lot

Answer 1

It is

$$a_{n+1} = \frac{2a_n}{1+a_n} = \frac{2*\frac{2^{n+1}}{2^{n+1}-1}}{1+\frac{2^{n+1}}{2^{n+1}-1}} = \frac{2^{n+2}}{2^{n+1}-1+2^{n+1}} = \frac{2^{(n+1)+1}}{2^{(n+1)+1}-1}$$

where the hypothesis is used in the second step.

Answer 2

What if you tried to show that $a_n$ is montonous, and bounded ?. Hint : what is the sign of $a_n$ ? of $a_n - a_{n+1}$ ? With that, you should be able to conclude.

Answer 3

I am sure that you will make "stand up" your recursive proof.

I propose here a different approach with a kind of geometrical flavour. Let us take the inverse of both sides of the defining relationship:

$$\frac{1}{a_{n+1}}=\frac{1}{2a_{n}}+\frac{1}{2} \tag{1}$$

Setting $b_n=\frac{1}{a_{n}}$, (1) becomes:

$$b_{n+1}=\frac{1}{2}(b_n+1)\tag{2}$$

(2) means that at each new step, the new value is in the middle of the former value and $1$, therefore, at each step, you divide by two the distance to $1$, explaining that: $b_n=1-\frac{1}{2^{n+1}}$ with $b_0=\frac12$ tending to $1$.

Now, think to invert $b_n$ to retrieve $a_n$ and its "reciprocal" behavior.