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

In summary, using a straightforward induction approach, it can be shown that ##x_{n+1}>x_n \because x_n + \sqrt{x_n} > x_n## for all values of n, starting from n=1. Additionally, it can be shown that ##\frac{(n+1)^2}{9} < \frac{n^2}{9} + \sqrt{\frac{n^2}{9}} \leq x_n + \sqrt{x_n} = x_{n+1}##, further supporting the original statement.
  • #1
yucheng
231
57
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
 
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  • #2
Say [tex]y_n=x_n−x_{n−1}[/tex] for n>1, y1=1 the relation is written as
[tex]y_{n+1}^2=x_n=\sum_{i=1}^n y_i [/tex]
For n=1 the statement stands,i.e.
[tex]x_1=1>\frac{1^2}{9}[/tex]
If it stand for n
[tex]y_{n+1}^2=x_n > \frac{n^2}{9}[/tex]
[tex]y_{n+1} > \frac{n}{3}[/tex]
Now you can show it stands for n+1.
 
Last edited:
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  • #3
anuttarasammyak said:
Say [tex]y_n=x_n−x_{n−1}[/tex] for n>1, y1=1 the relation is written as
[tex]y_{n+1}^2=x_n=\sum_{i=1}^n y_i [/tex]
For n=1 the statement stands,i.e.
[tex]x_1=1>\frac{1^2}{9}[/tex]
If it stand for n
[tex]y_{n+1}^2=x_n > \frac{n^2}{9}[/tex]
[tex]y_{n+1} > \frac{n}{3}[/tex]
You can show it stand for n+1.

did you mean this (I changed all [tex] to ##\#\###)

Say ##y_n=x_n−x_{n−1}## for n>1, y1=1 the relation is written as
##y_{n+1}^2=\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=\sum_{i=1}^n y_i > \frac{n^2}{9}##
##y_{n+1} > \frac{n}{3}##
You can show it stand for ##n+1##.
 
  • #4
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)
 
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  • #5
@Office_Shredder
I guess we always forget the most straightforward and basic methods :doh:thanks!
 
  • #6
yucheng said:
@Office_Shredder
I guess we always forget the most straightforward and basic methods :doh:thanks!
So, did you find the answer?Tell us about it if you want.
 
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  • #7
anuttarasammyak said:
Say [tex]y_n=x_n−x_{n−1}[/tex] for n>1, y1=1 the relation is written as
[tex]y_{n+1}^2=x_n=\sum_{i=1}^n y_i [/tex]
For n=1 the statement stands,i.e.
[tex]x_1=1>\frac{1^2}{9}[/tex]
If it stand for n
[tex]y_{n+1}^2=x_n > \frac{n^2}{9}[/tex]
[tex]y_{n+1} > \frac{n}{3}[/tex]
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}##
 
  • #8
##y_1=1## to be add and the final term be n not n+1.
 
  • #9
anuttarasammyak said:
##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.
 
  • #10
Office_Shredder said:
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)

infinitely small said:
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)
 

What is a sequence in mathematics?

A sequence in mathematics is a list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term, and the position of the term in the sequence is called its index. Sequences are commonly represented using the notation {an}, where "a" represents the term and "n" represents the index.

What is a lower bound for a sequence?

A lower bound for a sequence is the smallest number that is greater than or equal to all the terms in the sequence. In other words, it is the minimum value that the terms in the sequence can take. A sequence can have multiple lower bounds, but there is always a unique greatest lower bound, also known as the infimum.

How do you prove the lower bound for a sequence?

To prove the lower bound for a sequence, you must show that the lower bound is greater than or equal to all the terms in the sequence. This can be done using a mathematical proof, which involves using logical reasoning and mathematical properties to demonstrate that the lower bound is indeed the smallest possible value for the sequence.

Why is proving the lower bound important?

Proving the lower bound for a sequence is important because it helps us understand the behavior and properties of the sequence. It also allows us to make conclusions about the sequence, such as whether it is bounded or unbounded, convergent or divergent, and whether it has a limit or not.

What are some common techniques used to prove the lower bound for a sequence?

There are several techniques that can be used to prove the lower bound for a sequence, including the induction method, the contradiction method, and the direct method. These techniques involve using different approaches to show that the lower bound is greater than or equal to all the terms in the sequence.

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