Part a) Easy to note , that $2 \geq x_{2k-1}>1,0<x_{2k}<1$ Let $x_{2k-1}=y_{2k-1}+1,x_{2k}=1-y_{2k}, 0<y_k \leq 1$ Then $y_{2k}=1-\sqrt{y_{2k-1}+9}+\sqrt{y_{2k-1}+4}<\frac{y_{2k-1}}{2}$ $y_{2k+1}=\sqrt{9-y_{2k}}-\sqrt{4-y_{2k}}-1<\frac{y_{2k}}{2}$ So $(y_{k})$ is decreased and bounded so has limit.Easy to find, that $y_k \to 0$ and so $x \to 1$

gausskarl wrote: The sequence $(x_n)$ is defined as follows: $$x_1=2,\, x_{n+1}=\sqrt{x_n+8}-\sqrt{x_n+3}$$for all $n\geq 1$. a. Prove that $(x_n)$ has a finite limit and find that limit. b. For every $n\geq 1$, prove that $$n\leq x_1+x_2+\dots +x_n\leq n+1.$$ For part (b), I use 2 resuls: (i) $|x_k-1|\le \frac{1}{4^{k-1}},k\ge 2$, and (ii) $x_{2k}+x_{2k+1}\le 2, k\ge 1.$

We will prove that: $A=x_1+...+x_n \ge n$. Using Cauchy-Schwarz and AM-GM, we have: $$A+2 > A+x_{n+1} =x_1+x_2+...x_{n+1}=2+5\sum_{k=1}^{n}\frac{1}{\sqrt{x_k+8}+\sqrt{x_k+3}} \ge 2+\frac{5n^2}{\sum_{k=1}^{n}\sqrt{x_k+8}+\sum_{k=1}^{n}\sqrt{x_k+3}}$$$$\Rightarrow A>\frac{5n^2}{\sum_{k=1}^{n}\sqrt{x_k+8}+\sum_{k=1}^{n}\sqrt{x_k+3}}\ge \frac{5n^2}{\sqrt{n(A+8n)}+\sqrt{n(A+3n)}} \ge \frac{5n^2}{\frac{17n+A}{6}+\frac{7n+A}{4}}$$$$\Rightarrow (A-n)(A+12n)>0 \Leftrightarrow A>n$$

Very intelligent solution!

RagvaloD wrote: Part a) Easy to note , that $2 \geq x_{2k-1}>1,0<x_{2k}<1$ Let $x_{2k-1}=y_{2k-1}+1,x_{2k}=1-y_{2k}, 0<y_k \leq 1$ Then $y_{2k}=1-\sqrt{y_{2k-1}+9}+\sqrt{y_{2k-1}+4}<\frac{y_{2k-1}}{2}$ $y_{2k+1}=\sqrt{9-y_{2k}}-\sqrt{4-y_{2k}}-1<\frac{y_{2k}}{2}$ So $(y_{k})$ is decreased and bounded so has limit.Easy to find, that $y_k \to 0$ and so $x \to 1$ a very nice solution but how can you finh out that $(y_{k})$ ? please sho meh dei wae

You use Weierstrass theorem to get the limit equation.

Here is my solution for part a Solution It's easy to see that: $x_n > 0, \forall n \in \mathbb{N}$ We have: $|x_{n + 1} - 1| = |\sqrt{x_n + 8} - \sqrt{x_n + 3} - 1| = |\sqrt{x_n + 8} - 3 - \sqrt{x_n + 3} + 2|$ $=$ $|x_n - 1| . \left|\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right| < |x_n - 1| . \left|\dfrac{1}{\sqrt{x_n + 8} + 3} + \dfrac{1}{\sqrt{x_n + 3} + 2}\right| < \left(\dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{2 + \sqrt{3}}\right)|x_n - 1|$ Let $q = \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{2 + \sqrt{3}}$ then $0 < q < \dfrac{1}{3} + \dfrac{1}{2} < 1$ So: $|x_{n + 1} - 1| < q |x_n - 1| < q^2 |x_{n - 1} - 1| < ... < q^n |x_1 - 1| = q^n$ Hence: $\lim |x_n - 1| = 0$ or $\lim x_n = 1$

Here is my solution for part b Solution We have: $x_{n + 1} - 1 = (x_n - 1) \left(\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right)$ But: $\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2} < 0$ so: - If $x_n > 1$ then $x_{n + 1} < 1$ - If $x_n < 1$ then $x_{n + 1} > 1$ But: $x_1 = 2 > 1$ so: $x_2 < 1$ then: $x_3 > 1$ ... Hence: $x_{2k} < 1$ and $x_{2k + 1} > 1$ $(k \in \mathbb{N})$ We also have: $x_n + x_{n + 1} - 2 = (x_n - 1) \left(1 + \dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right)$ Combine with: $1 + \dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2} > 0$ so: - If $x_n > 1$ then $x_n + x_{n + 1} > 2$ - If $x_n < 1$ then $x_n + x_{n + 1} < 2$ But: $x_{2k} < 1$ and $x_{2k + 1} > 1$ so: $x_{2k} + x_{2k + 1} < 2$ and $x_{2k + 1} + x_{2k + 2} > 2$ Let $S_n = x_1 + x_2 + ... + x_n$ With $n = 2k (k \in \mathbb{N})$, we have: $S_{2k} = x_1 + x_2 + ... + x_{2k} = (x_1 + x_2) + (x_3 + x_4) + ... + (x_{2k - 1} + x_{2k}) > 2 + 2 + ... + 2 = 2k$ and $S_{2k} = x_1 + x_2 + ... + x_{2k} = x_1 + (x_2 + x_3) + (x_4 + x_ 5) + ... + (x_{2k - 2} + x_{2k - 1}) + x_{2k} < 2 + 2 + ... + 2 + 1 = 2k + 1$ So: $2k < S_{2k} < 2k + 1$ With $n = 2k + 1 (k \in \mathbb{N})$, we have: $S_{2k + 1} = x_1 + x_2 + ... + x_{2k + 1} = (x_1 + x_2) + (x_3 + x_4) + ... + (x_{2k - 1} + x_{2k}) + x_{2k + 1} > 2 + 2 + ... + 2 + 1 = 2k + 1$ and $S_{2k + 1} = x_1 + x_2 + ... + x_{2k + 1} = x_1 + (x_2 + x_3) + (x_4 + x_ 5) + ... + (x_{2k} + x_{2k + 1}) < 2 + 2 + ... + 2 + 2 = 2k + 2$ Then: $2k + 1 < S_{2k + 1} < 2k + 2$ Hence: $n < x_1 + x_ 2 + ... + x_n < n + 1, \forall n \in \mathbb{N}$

khanhnx wrote: Here is my solution for part a Solution It's easy to see that: $x_n > 0, \forall n \in \mathbb{N}$ We have: $|x_{n + 1} - 1| = |\sqrt{x_n + 8} - \sqrt{x_n + 3} - 1| = |\sqrt{x_n + 8} - 3 - \sqrt{x_n + 3} + 2|$ $=$ $|x_n - 1| . \left|\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right| < |x_n - 1| . \left|\dfrac{1}{\sqrt{x_n + 8} + 3} + \dfrac{1}{\sqrt{x_n + 3} + 2}\right| < \left(\dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{2 + \sqrt{3}}\right)|x_n - 1|$ Let $q = \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{2 + \sqrt{3}}$ then $0 < q < \dfrac{1}{3} + \dfrac{1}{2} < 1$ So: $|x_{n + 1} - 1| < q |x_n - 1| < q^2 |x_{n - 1} - 1| < ... < q^n |x_1 - 1| = q^n$ Hence: $\lim |x_n - 1| = 0$ or $\lim x_n = 1$ khanhnx wrote: Here is my solution for part b Solution We have: $x_{n + 1} - 1 = (x_n - 1) \left(\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right)$ But: $\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2} < 0$ so: - If $x_n > 1$ then $x_{n + 1} < 1$ - If $x_n < 1$ then $x_{n + 1} > 1$ But: $x_1 = 2 > 1$ so: $x_2 < 1$ then: $x_3 > 1$ ... Hence: $x_{2k} < 1$ and $x_{2k + 1} > 1$ $(k \in \mathbb{N})$ We also have: $x_n + x_{n + 1} - 2 = (x_n - 1) \left(1 + \dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right)$ Combine with: $1 + \dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2} > 0$ so: - If $x_n > 1$ then $x_n + x_{n + 1} > 2$ - If $x_n < 1$ then $x_n + x_{n + 1} < 2$ But: $x_{2k} < 1$ and $x_{2k + 1} > 1$ so: $x_{2k} + x_{2k + 1} < 2$ and $x_{2k + 1} + x_{2k + 2} > 2$ Let $S_n = x_1 + x_2 + ... + x_n$ With $n = 2k (k \in \mathbb{N})$, we have: $S_{2k} = x_1 + x_2 + ... + x_{2k} = (x_1 + x_2) + (x_3 + x_4) + ... + (x_{2k - 1} + x_{2k}) > 2 + 2 + ... + 2 = 2k$ and $S_{2k} = x_1 + x_2 + ... + x_{2k} = x_1 + (x_2 + x_3) + (x_4 + x_ 5) + ... + (x_{2k - 2} + x_{2k - 1}) + x_{2k} < 2 + 2 + ... + 2 + 1 = 2k + 1$ So: $2k < S_{2k} < 2k + 1$ With $n = 2k + 1 (k \in \mathbb{N})$, we have: $S_{2k + 1} = x_1 + x_2 + ... + x_{2k + 1} = (x_1 + x_2) + (x_3 + x_4) + ... + (x_{2k - 1} + x_{2k}) + x_{2k + 1} > 2 + 2 + ... + 2 + 1 = 2k + 1$ and $S_{2k + 1} = x_1 + x_2 + ... + x_{2k + 1} = x_1 + (x_2 + x_3) + (x_4 + x_ 5) + ... + (x_{2k} + x_{2k + 1}) < 2 + 2 + ... + 2 + 2 = 2k + 2$ Then: $2k + 1 < S_{2k + 1} < 2k + 2$ Hence: $n < x_1 + x_ 2 + ... + x_n < n + 1, \forall n \in \mathbb{N}$ Very intelligent solution!

teomihai wrote: khanhnx wrote: Here is my solution for part a Solution It's easy to see that: $x_n > 0, \forall n \in \mathbb{N}$ We have: $|x_{n + 1} - 1| = |\sqrt{x_n + 8} - \sqrt{x_n + 3} - 1| = |\sqrt{x_n + 8} - 3 - \sqrt{x_n + 3} + 2|$ $=$ $|x_n - 1| . \left|\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right| < |x_n - 1| . \left|\dfrac{1}{\sqrt{x_n + 8} + 3} + \dfrac{1}{\sqrt{x_n + 3} + 2}\right| < \left(\dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{2 + \sqrt{3}}\right)|x_n - 1|$ Let $q = \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{2 + \sqrt{3}}$ then $0 < q < \dfrac{1}{3} + \dfrac{1}{2} < 1$ So: $|x_{n + 1} - 1| < q |x_n - 1| < q^2 |x_{n - 1} - 1| < ... < q^n |x_1 - 1| = q^n$ Hence: $\lim |x_n - 1| = 0$ or $\lim x_n = 1$ khanhnx wrote: Here is my solution for part b Solution We have: $x_{n + 1} - 1 = (x_n - 1) \left(\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right)$ But: $\dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2} < 0$ so: - If $x_n > 1$ then $x_{n + 1} < 1$ - If $x_n < 1$ then $x_{n + 1} > 1$ But: $x_1 = 2 > 1$ so: $x_2 < 1$ then: $x_3 > 1$ ... Hence: $x_{2k} < 1$ and $x_{2k + 1} > 1$ $(k \in \mathbb{N})$ We also have: $x_n + x_{n + 1} - 2 = (x_n - 1) \left(1 + \dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2}\right)$ Combine with: $1 + \dfrac{1}{\sqrt{x_n + 8} + 3} - \dfrac{1}{\sqrt{x_n + 3} + 2} > 0$ so: - If $x_n > 1$ then $x_n + x_{n + 1} > 2$ - If $x_n < 1$ then $x_n + x_{n + 1} < 2$ But: $x_{2k} < 1$ and $x_{2k + 1} > 1$ so: $x_{2k} + x_{2k + 1} < 2$ and $x_{2k + 1} + x_{2k + 2} > 2$ Let $S_n = x_1 + x_2 + ... + x_n$ With $n = 2k (k \in \mathbb{N})$, we have: $S_{2k} = x_1 + x_2 + ... + x_{2k} = (x_1 + x_2) + (x_3 + x_4) + ... + (x_{2k - 1} + x_{2k}) > 2 + 2 + ... + 2 = 2k$ and $S_{2k} = x_1 + x_2 + ... + x_{2k} = x_1 + (x_2 + x_3) + (x_4 + x_ 5) + ... + (x_{2k - 2} + x_{2k - 1}) + x_{2k} < 2 + 2 + ... + 2 + 1 = 2k + 1$ So: $2k < S_{2k} < 2k + 1$ With $n = 2k + 1 (k \in \mathbb{N})$, we have: $S_{2k + 1} = x_1 + x_2 + ... + x_{2k + 1} = (x_1 + x_2) + (x_3 + x_4) + ... + (x_{2k - 1} + x_{2k}) + x_{2k + 1} > 2 + 2 + ... + 2 + 1 = 2k + 1$ and $S_{2k + 1} = x_1 + x_2 + ... + x_{2k + 1} = x_1 + (x_2 + x_3) + (x_4 + x_ 5) + ... + (x_{2k} + x_{2k + 1}) < 2 + 2 + ... + 2 + 2 = 2k + 2$ Then: $2k + 1 < S_{2k + 1} < 2k + 2$ Hence: $n < x_1 + x_ 2 + ... + x_n < n + 1, \forall n \in \mathbb{N}$ Very intelligent solution! Arent these a bit too overwhelming for a P1? Does anyone have a bit more intuitive solution ?

This is more of a sketch than a solution. Solution of part b. Claim 1. If $a_i>1$, then $a_{i+1}<1$ Proof. \begin{align*} x_{n+1}^2 &= 2x_n+11-2\sqrt{(x_n+3)(x_n+8)}\\ &< 2x_n+11-2\sqrt{(x_n+5)^2}=1 \end{align*}as $(a+3)(a+8)>(a+5)^2 \iff a>1$ We'll use strong induction. Base case is trivial. Assume that it's true for $n\in \{1,2,..,k\}$. 1. If $a_k>1$, then $a_{k+1}<1$ from Claim 1. So $a_1+a_2+...+a_{k+1}\leq k+2$. It can be easily proved that $a_k+a_{k+1}\ge 2$ implying $(a_1+..+a_{k-1})+a_k+a_{k+1} \ge k+1$. 2. If $a_k<1$, it can be easily proved that $a_{k+1}>1$, $a_k+a_{k+1}\leq 2$. So it follows. 3. $a_k=1$. Trivial.

I like how everyone uses "hence" and "thus".

Psyduck909 wrote: Arent these a bit too overwhelming for a P1? Does anyone have a bit more intuitive solution ? Here's a more intuitive solution (along with motivation): Define $y_n = x_n - 1$. Then $$y_1, = 2, ~ y_{n+1} = \sqrt{y_n + 9} - \sqrt{y_n + 4} - 1$$and we want to prove $$0 \le y_1 + y_2 + \cdots + y_n \le 1 \qquad \forall ~ n \ge 1$$Observe that \begin{align*} y_{n+1} &= \sqrt{y_n + 9} - \sqrt{y_n + 4} - 1 = (\sqrt{y_n + 9} - 3) - (\sqrt{y_n + 4} - 2)\\ &= \frac{y_n}{\sqrt{y_n + 9} + 3} - \frac{y_n}{\sqrt{y_n + 4} + 2} = y_n \left( \frac{1}{\sqrt{y_n + 9} + 3} - \frac{1}{\sqrt{y_n + 4} + 2} \right) \\ &= -y_n \left( \frac{1}{\sqrt{y_n + 4} + 2} - \frac{1}{\sqrt{y_n + 9} + 3} \right) \end{align*}Now define $$c_n = \frac{1}{\sqrt{y_n + 4} + 2} - \frac{1}{\sqrt{y_n + 9} + 3}$$Observe that $0 < c_n < 1 ~ \forall ~ n \ge 1$ and we also have $$y_{n+1} = -y_n \cdot c_n \qquad \forall ~ n \ge 1$$Hence, \begin{align*} & y_1 + y_2 + \cdots + y_n = (\underbrace{y_1 - y_2}_{\ge 0}) + (\underbrace{y_3 - y_4}_{\ge 0}) + \cdots \ge 0 \\ & y_1 + y_2 + \cdots + y_n = y_1 + (\underbrace{y_2 - y_3}_{\le 0}) + (\underbrace{y_4 - y_5}_{\le 0}) + \cdots \le y_1 = 1 \end{align*}Note that in the first, second case, one positive, negative (respectively) term possibly may not be paired, but of course that works good with the sign of the inequality. $\blacksquare$

gausskarl wrote: The sequence $(x_n)$ is defined as follows: $$x_1=2,\, x_{n+1}=\sqrt{x_n+8}-\sqrt{x_n+3}$$for all $n\geq 1$. a. Prove that $(x_n)$ has a finite limit and find that limit. For if the limit exists and is equal to $k$ then the given recursion implies that $k=\sqrt{k+8}-\sqrt{k+3}$ for which the only solution is $k=1$. From $x_n\to 1$ we show that $$|x_{n+1}-1|<\frac{1}{10}|x_n-1|$$This can be demonstrated by graphing. Then $|x_n-1|$ is squeezed to $0$, so $x_n$ is squeezed to $1$.

Essentially we just analyze the sequence. Let $y_n = x_n - 1$ for each $n$. Claim. $\{y_n\}$ is an alternating series with its terms decreasing in magnitude. Proof. For the first part, suppose that $y_n > 0$. Then as $f(t) = \sqrt{t+9} - \sqrt{t+4} - 1$ is decreasing (say by taking a derivative), it follows that $y_{n+1} = f(y_n) < f(1) = 0$. For the second part, observe that the inequality \[\left|\sqrt{t+9} - \sqrt{t+4} - 1\right| < |t|\]holds for all $t$ by checking $t$ positive and $t$ negative cases. $\blacksquare$ Now both parts immediately follow; the second part follows because for $S_n$ the sum of the $y_i$, we have $S_n \in [S_1, S_2] \subset \left(\sqrt{10}-\sqrt 5, 1\right)$, directly implying the result.