- #1
DivGradCurl
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Folks,
This is the solution I have for a problem in my textbook regarding sequences. I just need to know whether I have the right idea in mind.
Thank you very much!
We can use an analogus function to show that the sequence given by
[tex] a_{n+1} = \sqrt{2+a_n} \quad a_1 = \sqrt{2} [/tex]
is increasing. Here it goes
[tex] y = \sqrt{2+x} = \left( 2+x \right) ^{\frac{1}{2}} [/tex]
[tex] \frac{dy}{dx}=\frac{1}{2\sqrt{2+x}}>0 \Longrightarrow a_{n+1}>a_n [/tex]
This is the solution I have for a problem in my textbook regarding sequences. I just need to know whether I have the right idea in mind.
Thank you very much!
We can use an analogus function to show that the sequence given by
[tex] a_{n+1} = \sqrt{2+a_n} \quad a_1 = \sqrt{2} [/tex]
is increasing. Here it goes
[tex] y = \sqrt{2+x} = \left( 2+x \right) ^{\frac{1}{2}} [/tex]
[tex] \frac{dy}{dx}=\frac{1}{2\sqrt{2+x}}>0 \Longrightarrow a_{n+1}>a_n [/tex]
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