What is the limit of the recurrence relation for g_n?

Marian123
Messages
1
Reaction score
0
Hi all

Suppose that , a_{n+1}=a_n^2-2 and g_n=\frac{a_1a_2...a_n}{a_{n+1}}.
Evaluate \lim_{n\rightarrow \infty } g_n.

I have seen some information in this link. Besides, the sequence gn seems as a good rational approximation for \sqrt5. I know that the answer is 1, But I can't find any nice solution. Any hint is strongly appreciated.
 
Mathematics news on Phys.org
I'm not sure the limit exists in general ...

If a_1=0 then g_n=0 for all n, so the limit is zero. If a_1=1, then g is an alternating sequence and the limit does not exist. If a_1=2, then g_n=2^{n-1} which diverges. In fact, after playing for a few minutes, I can only get g to either converge to zero or not converge at all.

jason
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top