Let u2>u1>0 and also let un+1=√unun−1 for all n≥2. Then prove that the sequence {un} converges.
For this, use of only the following results is permissible,
For two sequence {xn} and {yn}
limn→∞(xn+yn)=limn→∞xn+limn→∞yn
limn→∞(xn⋅yn)=(limn→∞xn)⋅(limn→∞yn)
provided they exists.
- For a sequence {zn}, the limit limn→∞zn doesn't exist is equivalent to saying that, ∃ε>0∣|zn−l|≥ε ∀l∈R∧∀n≥n0(∈N)
I have been able to prove that limn→∞(unu2n+1)=u1u22. Now one can conclude from 1 and 2 that the limit must be 3√u1u22 but that happens only if we can prove that the limits exist. And this is exactly where I am stuck. Using only the three mentioned results I can't prove that. Any help will be appreciated.
Answer
This is similar to what you was trying, but if it helps... (apologies for my poor English)
Supposing you are allowed to use induction, then
u2n+1=unun−1⟹u2n+1un=u2nun−1=...=u22u1
Making u22u1=a you have un+1=√a/un=4√aun−1 and again by induction un+1=asnucni
i∈{1,2}, sn is a partial sum of a geometric series, and cn a power of 14, all depending on the parity of n.
Now you can use the result number (3) to prove that each element on the right member of the last equality has a limit (I think this is an easier problem than the initial), and then use result (2) to conclude.
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