Mathematica Can You Prove \(x_n < x_{n+1}\) Using Induction?

AI Thread Summary
The discussion centers on proving that the sequence defined by x_1 = √2 and x_{n+1} = √(2 + x_n) satisfies the inequality x_n < x_{n+1} using mathematical induction. One participant suggests that induction is necessary to establish the inequality, while another proposes an alternative approach by analyzing the expression under the radical. The key steps in the induction proof involve showing that x_1 < x_2, assuming x_k < x_{k+1}, and then demonstrating that this leads to x_{k+1} < x_{k+2}. The increasing nature of the function f(a) = a^(1/2) is also highlighted as crucial to the argument. The discussion concludes with a clearer understanding of the proof process for the sequence.
MrBailey
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Hi, all.
I'm working on some proof by induction problems. While I understand the concept, this one threw me for a loop.
Let x_1=\sqrt{2} and x_{n+1}=\sqrt{2+x_n}
Show that x_n &lt; x_{n+1}
I'd greatly appreciate help with this.
Thanks,
bailey
 
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sure you need to use induction? i would show that the stuff under the radical for x_{n+1}&gt; x_n we know this because x_n&gt;0. and 2 plus some other positive number will always be greater than two, and therefore the sq rt of that sum will be greater eh?
 
Uh, you certainly need induction. If xn = 98, then xn+1 = (2 + 98)1/2 = 1001/2 = 10 < 98 = xn.

Show that x1 < x2
Assume that xk < xk+1
Use this to prove that xk+1 < xk+2
Write out xk+1 and xk+2 in terms of xk. Then there xk+1 < xk+2 will follow immediately from xk < xk+1 as long as you know that the function f defined by f(a) = a1/2 is an increasing function.
 
thanks...much clearer now

Bailey
 

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