# Order of convergence of sequence

1. Sep 14, 2011

### Damidami

1. The problem statement, all variables and given/known data

I have to find the order of convergence of the following sequence

$b_n = \left( \frac{5}{6} \right)^{n^2}$

I have numerically tested that it has to be a real number between 1 and 2, but I can't find it exactly.

I also have this doubt: does every sequence have an order of convergence? (a real number $p$ such that $| b_{n+1}/b_{n}^p|$ converges to a non-zero constant )

2. Relevant equations

3. The attempt at a solution

Here http://translate.google.com/transla...xactas.wordpress.com/2011/08/29/tp-1-ej-1-3/" are my futile attempts at solving it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Apr 26, 2017
2. Sep 14, 2011

### grief

It seems to me that there is no order of convergence in this case, according to the definition you gave. Try to plug the formula into the definition of order of convergence.

$\frac{5}{6}^{{(n+1)}^2} / {(\frac{5}{6}^{n^2})}^p = \frac{5}{6}^{{(n+1)}^2-pn^2}$. If p<=1, then the exponent tends to infinity, so the whole expression tends to 0. If p>1, then the exponent tends to -infinity and the whole expression tends to infinity.