Define a real sequence \({a}_{n}\) by \({a}_{1}=\sqrt{2}\) and \({a}_{n+1}=\sqrt{2+\sqrt{{a}_{n}}}\) for n>=2

(1) show the sequence \({a}_{n}\) is bounded

(2) Explain why the sequence \({a}_{n}\) converges.

You can use the following definition to proof it.

Suppose (\({a}_{n}\)) is a sequence and L ∈ C such that for all \(\varepsilon \) > 0 there is an integer N such that for all n ≥ N, we have |\({a}_{n}\)− L| < \(\varepsilon \). Then the sequence (an) is convergent and L is its limit; in symbols we write \({lim}_{n\rightarrow infintiy}\)\({a}_{n}\) = L . If no such L exists then the sequence (\({a}_{n}\)) is divergent.

(3)Find the limit of the sequence \({a}_{n}\)

I can use algebra to solve this , but since this is complex analysis problem, you might use previous definition to solve this. I am also seeking general formula for \({a}_{n}\) in term of n.

fiora
Nov 20, 2018

#1**0 **

what do you mean by "L is a member of C"? I think it's supposed to be "L is a real number" if the sequence is a sequence of real numbers.

also I don't understand how this is a complex analysis problem.

Guest Nov 20, 2018

#2**0 **

Because this question show up on my complex analysis test, and the test covered the topic of complex sequence and series.But why is this on my test? This is also my question for my wired professor. Complex number include real number(complex numberz=x+i*y, for real x and y, if y=0 then z is real number), so the definition can also apply to (2). The definition is indeed in my complex analysis textbook, and normally I can use this defintion to proof the sequence converge or diverge. But in this case, I guess I can show the sequence is bounded and then by some theorem/corrollay to show thatthe sequence is converge. If don't know solve this with any kind of math, just leave this alone and don't bother the defintion stated in my original question. Talk is cheap, show me the math(code).--Linus Trovalds

fiora
Nov 20, 2018