- #1
yucheng
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- 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
$$
\begin{align*}
x_{n+1} &= x_n+ \sqrt{x_n} \\
&= x_1 + \sqrt{x_1} + \sqrt{x_2} + \cdots \sqrt{x_n} \\
&>n+1
\end{align*}
$$
##\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