Prove the lower bound for a sequence (Buck, Advanced Calculus)

[yucheng]

Homework Statement

Define a sequence ${x_n}$ by $$x_1=1, x_{n+1} = x_n + \sqrt{x_n}$$

Show that $\forall n\geq 1: x_n \geq \frac{n^2}{9}$

(Buck, Advanced Calculus, section 1.6, Exercise 16)

Relevant Equations

N/A

Clearly, $x_{n+1}>x_n \because x_n + \sqrt{x_n} > x_n$
$$
\begin{align*}
x_{n+1} &= x_n+ \sqrt{x_n} \\
&= x_1 + \sqrt{x_1} + \sqrt{x_2} + \cdots \sqrt{x_n} \\
&>n+1
\end{align*}
$$
$\because \sqrt{x_n}>\sqrt{x_1}=1$

In fact, $$x_{n+1} > 1+ \sqrt{1} + \sqrt{2}+ \sqrt{3} + \cdots \sqrt{n}$$

So... Might this help? Closed form for sum of square roots

 

[anuttarasammyak]

Say y_n=x_n−x_{n−1} for n>1, y1=1 the relation is written as
y_{n+1}^2=x_n=\sum_{i=1}^n y_i
For n=1 the statement stands,i.e.
x_1=1>\frac{1^2}{9}
If it stand for n
y_{n+1}^2=x_n > \frac{n^2}{9}
y_{n+1} > \frac{n}{3}
Now you can show it stands for n+1.

 

[yucheng]

Say y_n=x_n−x_{n−1} for n>1, y1=1 the relation is written as
y_{n+1}^2=x_n=\sum_{i=1}^n y_i
For n=1 the statement stands,i.e.
x_1=1>\frac{1^2}{9}
If it stand for n
y_{n+1}^2=x_n > \frac{n^2}{9}
y_{n+1} > \frac{n}{3}
You can show it stand for n+1.

did you mean this (I changed all to $\#\$#)<br /> <br /> Say $y_n=x_n−x_{n−1}$ for n&gt;1, y1=1 the relation is written as<br /> $y_{n+1}^2=\sum_{i=1}^n y_i$<br /> For n=1 the statement stands,i.e.<br /> $x_1=1&gt;\frac{1^2}{9}$<br /> If it stand for n<br /> $y_{n+1}^2=\sum_{i=1}^n y_i &gt; \frac{n^2}{9}$<br /> $y_{n+1} &gt; \frac{n}{3}$<br /> You can show it stand for $n+1$.

 

[Office_Shredder]

I think you can just do this by mostly straightforward induction. In particular notice that $\frac{(n+1)^2}{9} < \frac{n^2}{9} + \sqrt{\frac{n^2}{9}},$ at least if n is big enough (proof of this left to the reader)

 

[yucheng]

@Office_Shredder
I guess we always forget the most straightforward and basic methods thanks!

 

[trees and plants]

@Office_Shredder
I guess we always forget the most straightforward and basic methods thanks!

So, did you find the answer?Tell us about it if you want.

 

[trees and plants]

Say y_n=x_n−x_{n−1} for n>1, y1=1 the relation is written as
y_{n+1}^2=x_n=\sum_{i=1}^n y_i
For n=1 the statement stands,i.e.
x_1=1&gt;\frac{1^2}{9}
If it stand for n
y_{n+1}^2=x_n &gt; \frac{n^2}{9}
y_{n+1} &gt; \frac{n}{3}
Now you can show it stands for n+1.

This is an different approach that i took to solve it. Does the equation $x_n=\sum_{i=1}^n y_i$ hold? How did you show it? Because if you try $\sum_{i=1}^n y_i=y_1+y_2+...+y_n=(x_1-x_0)+(x_2-x_1)+(x_3-x_2)+(x_4-x_3)+(x_5-x_4)+$ $...+(x_n-x_{n-1})+(x_{n+1}-x_n)=-x_0+x_{n+1}=-1+x_{n+1}$

 

[anuttarasammyak]

$y_1=1$ to be add and the final term be n not n+1.

 

[trees and plants]

$y_1=1$ to be add and the final term be n not n+1.

You are correct, I saw it now. Sorry for that.

 

[yucheng]

I think you can just do this by mostly straightforward induction. In particular notice that $\frac{(n+1)^2}{9} < \frac{n^2}{9} + \sqrt{\frac{n^2}{9}},$ at least if n is big enough (proof of this left to the reader)

So, did you find the answer?Tell us about it if you want.

Yup. Base case: $x_1=1 \geq 1/9$ Assume inductively that $x_n \geq \frac{x^2}{9}$, then $\frac{(n+1)^2}{9} < \frac{n^2}{9} + \sqrt{\frac{n^2}{9}} \leq x_n + \sqrt{x_n} = x_{n+1}$

P.S. $\frac{3n}{9} > \frac{2n+1}{9} \because n>1$ (except base case)