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

Prove that the series $c_{1} = 1$, $c_{(n+1)}= 4/(1 + 5c_{n})$, $n \geq 1$ is convergent and also locate its restriction.

Ok so yet I've exercised a number of things.
$c_1 = 1$
$c_2 = 2/3$
$c_3 = 12/13$
$c_4 = 52/73$

So the weird $c_n$ are lowering and also the also $c_n$ are raising. With ease, it is clear the both series for weird and also also $c_n$ are decreasing/increasing much less and also much less. Consequently it feels like the series might merge to some restriction $L$.

If the series has a restriction, allow $L=\underset{n\rightarrow \infty }{\lim }a_{n}.$ Then $L = 1/(1+5L).$ So we generate $L = 4/5$ and also $L = -1$. Yet given that the even series is raising and also > 0, after that $L$ has to be $4/5$.

Ok, below I am stuck. I'm not exactly sure just how to proceed and also show that the series merges to this restriction (I attempted making use of the definition of the restriction yet I really did not take care of) and also as well as not exactly sure concerning the different series just how I would certainly deal with revealing their restrictions.

A couple of notes: I remain in 2nd year calculus. This is an incentive inquiry, yet I appreciate the obstacle and also would certainly enjoy the added marks. Keep in mind: Once once more I apologize I do not recognize just how to make use of the HTML code to make it wonderful.

2019-05-18 22:39:44
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Answers: 5

Here is one means to confirm it: allow $f(x) = 4/(1+5x)$. Claim $|x-4/5| \le C$ for some constant $C$. Can you locate $C$ and also some constant $0 \le k < 1$ to make sure that if $|x-4/5| \le C$, after that $|f(x)-4/5| \le k|x-4/5|$?

If you do this, after that you can iterate to get $|f^j(x)-4/5| \le k^j |x-4/5|$, for all $j$, therefore if you make $j$ huge adequate after that you can get $f^j(x)$ as near $4/5$ as you such as.

2019-05-21 07:18:12

As revealed by the various other solutions, there are a couple of wonderful means to approach this trouble.

You can focus just on $C_{2n-1},$ claim, given that if you develop that $C_{2n-1}$ often tends to a restriction you instantly toenail $C_{2n}$ too, due to the fact that

$$C_{2n} = \frac{4}{1+5C_{2n-1}} \textrm { so } \lim C_{2n} = \lim \frac{4}{1+5C_{2n-1}}.$$

Now it is very easy to show $C_{2n-1} > 4/5$ therefore expand $$(5C_{2n-1}-4)(C_{2n-1}+1)>0$$

and also adjust (add $20C_{2n-1}$ to both sides and also take the 4 to the RHS) to get

$$ C_{2n-1} > \frac{4+20C_{2n-1}}{21+5C_{2n-1}} = C_{2n+1},$$

and also given that $C_{2n-1}$ is bounded listed below by 4/ 5 the outcome adheres to quickly.

2019-05-21 07:04:44

Well done! Your monitorings are proper and also can be finished to offer an evidence that the series is convergent to $\frac{4}{5}$.

The challenging component in confirming your monitorings it to confirm that the odd/even below - series are monotonic and also suitably bounded (what do I suggest by this?).

Below is a tip:

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


Try writing $\displaystyle c_{n+2}$ in regards to $\displaystyle c_{n}$ and also see if that aids you confirm the above bounds (and also as a next action, the monotonicity).

(For a less complex evidence of the above bound, you can additionally attempt revealing that if $c_{n} \ge 4/5$ after that $c_{n+1} \le 4/5$ and also in a similar way if $c_n \le 4/5$ after that $c_{n+1} \ge 4/5$).

2019-05-21 06:59:08

So $c_{1} = 1$ and also $c_{n+1} = 4/(1+5c_{n})$ for $n \geq 1$. Allow $C(x) = \sum_{n \geq 1} c_{n}x^n$. After that possibly attempt to share $\sum_{n \geq 1} c_{n+1}x^n$ and also $\sum_{n \geq 1} \frac{4x^n}{1+5c_{n}}$ in regards to $C(x)$ to get a shut kind.

2019-05-21 06:58:19

Here is an extra specific variation of my tip. I claimed it would certainly be less complicated to show that $d_n = c_n - \frac{4}{5}$ often tends to absolutely no as $n \to \infty$. This offers $d_1 = \frac{1}{5}$ and also the reappearance

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


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

Can you see what to do from below?

Edit: Some extra tips. If the $1 + d_n$ in the were simply a $1$, we would certainly be done due to the fact that after that $|d_n|$ would certainly decrease greatly at the very least as quickly as $\left( \frac{4}{5} \right)^n$. Yet it is not. Nonetheless, $d_n$ is tiny, so the needs to be close sufficient to $1$ that this argument need to still experience. Extra specifically, you simply require to locate a constant $0 < c < \frac{1}{5}$ such that you can confirm that $|d_n| \le c$, claim for $n \ge 2$. (Maybe by induction.) From there it will certainly adhere to that $|d_n|$ in fact decomposes greatly at the very least as quickly as $\left( \frac{4}{5(1 - c)} \right)^n$.

This is an instance of a basic strategy called "bootstrapping," where you make use of weak price quotes along with relationships that a series satisfies to get more powerful price quotes.

2019-05-21 06:34:23