Prove that the sequence$ c_1 = 1$, $c_{n+1} = 4/(1 + 5c_n) $ , $ n \geq 1$ is convergent and find its limit

Solution 1:

Here is a more explicit version of my hint. I said it would be easier to show that $d_n = c_n - \frac{4}{5}$ tends to zero as $n \to \infty$. This gives $d_1 = \frac{1}{5}$ and the recurrence

$$d_{n+1} + \frac{4}{5} = \frac{4}{1 + 5(d_n + \frac{4}{5})} = \frac{4}{5} \frac{1}{1 + d_n}$$

hence

$$d_{n+1} = - \frac{4}{5} \frac{d_n}{1 + d_n}.$$

Can you see what to do from here?

Edit: Some more hints. If the $1 + d_n$ in the denominator were just a $1$, we would be done because then $|d_n|$ would decrease exponentially at least as fast as $\left( \frac{4}{5} \right)^n$. But it's not. However, $d_n$ is small, so the denominator should be close enough to $1$ that this argument should still go through. More precisely, you just need to find a constant $0 < c < \frac{1}{5}$ such that you can prove that $|d_n| \le c$, say for $n \ge 2$. (Maybe by induction.) From there it will follow that $|d_n|$ actually decays exponentially at least as fast as $\left( \frac{4}{5(1 - c)} \right)^n$.

This is an example of a general technique called "bootstrapping," where you use weak estimates together with relations that a sequence satisfies to get stronger estimates.

Solution 2:

Well done! Your observations are correct and can be completed to give a proof that the sequence is convergent to $\frac{4}{5}$.

The tough part in proving your observations it to prove that the odd/even sub-sequences are monotonic and appropriately bounded (what do I mean by this?).

Here is a hint:

Show that $$ c_{2n-1} \ge \frac{4}{5} \ge c_{2n} \ \ \forall n \ge 1$$

Hint2:

Try writing $\displaystyle c_{n+2}$ in terms of $\displaystyle c_{n}$ and see if that helps you prove the above bounds (and as a next step, the monotonicity).

(For a simpler proof of the above bound, you can also try showing that if $c_{n} \ge 4/5$ then $c_{n+1} \le 4/5$ and similarly if $c_n \le 4/5$ then $c_{n+1} \ge 4/5$).