Prove that the sequence: $a_1 = 1, a_{n+1} =\sqrt{c+da_n}$ (when the real numbers $c, d > 1$) is converging and find it's limit

Question

I have a summarized solution but it's starts with proving that the sequence is bounded from above by c+d. How can I know that this sequence is bounded by c+d? I understand the proof by induction but how do I actually realize that fact?

I know how to prove that the sequence is monotonically increasing and to find it's limit after we've established that the sequence is converging.

Answer 1

Hint: If you already managed to prove that the sequence is increasing, you have that the sequence is either divergent to $+\infty$ or converging to some limit $l>0$. In the second case, such a number must fulfill $l=\sqrt{c+dl}$, hence: $$ l=\frac{d+\sqrt{4c+d^2}}{2}. $$ On the other hand, $\{a_n\}_{n\geq 1}$ is bounded: $$ a_{n+1}^2 \leq (c+d)\,a_n\quad\Longrightarrow\quad \left(\frac{a_{n+1}^2}{a_n}\right)\cdot\left(\frac{a_{n}^2}{a_{n-1}}\right)^{1/2}\cdot\ldots\cdot\left(\frac{a_{2}^2}{a_1}\right)^{2/2^n}\leq (c+d)^2$$ from which it follows that $a_n\leq(c+d)$ for any $n\geq 1$.

Answer 2

If you want an upper bound $M$ for the sequence $a_n$, then you're going to want:

• $a_1=1\le M$
• $a_n\le M\implies a_{n+1}\le M$
  • Now, if all we know about $a_n$ is that it's $\le M$, then the best bound we can have for $a_{n+1}=\sqrt{c+da_n}\le\sqrt{c+dM}$. For this to guarantee $a_{n+1}\le M$, we need $\sqrt{c+dM}\le M$, i.e. $M^2-dM-c\ge0$. If we call the negative root of this equation $-s$, and the positive root $r$, this becomes $(M+s)(M-r)\ge0$, which leads to $M\ge r$ (excluding the case of negative bounds because we know $a_n\ge1>0$). So, the tightest upper bound we can establish for $a_n$ is $r$ (which we probably suspect to be its limit).

To confirm that our sequence is monotonically increasing, we note that for $a_n \ge 0$, $a_{n+1}\ge a_n\iff a_{n+1}^2\ge a_n^2\iff c+da_n\ge a_n^2$

$$\iff a_n^2-da_n-c\le0\iff (a_n+s)(a_n-r)\le0$$

Now, we know that $0\le a_n \le r$, so this sequence is increasing for all $n$ and bounded above by $r$, hence convergent.

Finally, denoting the limit of $a_n$ by $L$, we see that $a_{n+1}$ will have the same limit, and standard observations tell us that $L=\sqrt{c+dL}$, which has solutions $L=-s,r$. Noting that $a_n\ge0$ for all $n$ gives that the limit must be $r$, and we are done.

Answer 3

We can apply $a_n \le d+2c+\frac{c^2-c}{d}$ also to show that $a_n$ is bounded, instead to show $a_n \le d+c$. The point is that we can still apply monotone convergence theorem.

For $n=1$, $a_1=1 < d+2c+\frac{c^2-c}{d}$. Then suppose that $a_n < d+2c+\frac{c^2-c}{d}$ for some $n$. Then \begin{align} a_{n+1}&=\sqrt{c+da_n}\\ &\le\sqrt{c+d\left(d+2c+\frac{c^2-c}{d}\right)}\\ &=\sqrt{c^2+2cd+d^2}\\ &=c+d\\ &\le c+d+\left(c+\frac{c^2-c}{d}\right)\\ &=d+2c+\frac{c^2-c}{d}. \end{align} By mathematical induction, the inequality holds for all positive integer $n$.

Answer 4

Better idea: the equation $a = \sqrt{c+da}$ has a solution $$a = a_0 := \frac{d+\sqrt{4c+d^2}}2>d>1$$ and for $a\in[1,a_0]$: $a\le\sqrt{c+da}\le a_0$, i.e., $$a_n\le a_{n+1}\le a_0$$

Answer 5

If a sequence is increasing and bounded, then it converges to the supremum of the values. If it is increasing and unbounded, then it diverges to $\infty$.

So all you need is to prove the sequence is increasing and bounded.

Let's try seeing $a_1\le a_2,$ which means $1\le\sqrt{c+d}$, which is true because $1\le c+d$.

Now, suppose you know that $a_n\le a_{n+1}$; then the inequality $$ a_{n+1}\le a_{n+2} $$ is equivalent to $$ c+da_n\le c+da_{n+1} $$ which clearly follows from the induction hypothesis.

We just have to guess a bound. But we know what to look for: the sequence is increasing, so the limit must be an upper bound.

So, suppose the sequence is convergent to $a$; then $$ a=\sqrt{c+da} $$ so $$ a^2=c+da $$ and $$ a^2-da-c=0 $$ so $$ a=\frac{d+\sqrt{d^2+4c}}{2} $$ must be the limit and an upper bound for the sequence.

Note that $a>1$, so the base of the induction is established. Suppose that $a_n\le a$; then $a_{n+1}\le a$ is equivalent to $$ c+da_n\le \left(\frac{d+\sqrt{d^2+4c}}{2}\right)^2 $$ which becomes $$ 4c+4da_n\le d^2+d^2+4c+2d\sqrt{d^2+4c} $$ that is, after simplifying, $$ 2a_n\le d+\sqrt{d^2+4c} $$ which is precisely the induction hypothesis.

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If you want to show the sequence is bounded by $c+d$, you can do this way. First, $a_1\le c+d$ is true. Suppose $a_n\le c+d$; then $a_{n+1}\le c+d$ is equivalent to $$ c+da_n\le (c+d)^2 $$ that is, $$ a_n\le\frac{(c+d)^2-c}{d} $$ and you'll be done if you prove that $$ \frac{(c+d)^2-c}{d}\ge c+d $$ which becomes $$ (c+d)^2-cd-d^2-c\ge0 $$ or $$ c^2+d(c-1)\ge0 $$ which is true.

Answer 6

If the sequence defined recursively by $a_{n+1}:=\sqrt{c+da_n}$ converges then the limit $\alpha>0$ satisfies the equation $$\alpha^2=c+d\alpha\ ,$$ hence $$\alpha={1\over2}\bigl(d+\sqrt{d^2+4c}\bigr)>d\ .$$ Therefore let's look at the quantities $x_n:=a_n-\alpha$. From the given recurrence we get $$x_{n+1}=a_{n+1}-\alpha={\bigl(c+ d( x_n+\alpha)\bigr)-\alpha^2\over\sqrt{c+da_n}+\alpha}={d\over\sqrt{c+da_n}+\alpha}\>x_n\qquad(n\geq0)\ .$$ Since $$0<{d\over\sqrt{c+da_n}+\alpha}<{d\over 1+d}<1$$ it follows that $\lim_{n\to\infty} x_n=0$, hence $\lim_{n\to\infty} a_n=\alpha$ whenever $a_0\geq0$.