# How to find range of values for which a sequence converges?

Say you had the sequence $U_{n+1} = 2bU_n$ where $U_1 = 6$. How would you find the range of values of b for which the sequence converges?

• I will assume you really mean sequence and not the series $\sum_1^\infty U_n$. So $U_1=6$, $U_2=6(2b)$, $U_3=6(2b)^2$, $U_4=6(2b)^3$, and so on. Maybe this will help. Dec 2, 2015 at 19:25
• Definitely a sequence. Dec 2, 2015 at 19:27

Essentially, you have to find out for which values of x a sequence x^n converges. These are all values, which converge to zero (don't miss the negative ones!) and one more special case.

• Sorry, I don't see how that relates to my sequence. Like, I know it means I have to find the values that the sequence will be heading to a limiting value. Dec 2, 2015 at 19:40
• As Andre commented above, the value of the sequence is 6 (2b)^n with n->inf, so set x=2b
– Ctx
Dec 2, 2015 at 19:44
• I'm trying but this just isn't clicking with me where this is going. So 2b^n, and I have to find the values of b that make it converge. I would naturally be thinking fractions here, but that's all I'm really thinking. Dec 2, 2015 at 19:58
• E.g.: What does 1.001^n converge to?
– Ctx
Dec 2, 2015 at 20:00
• 2? Eventually though it would go above 2. Would it be all the values when 2b < 1? Dec 2, 2015 at 20:07

The sequence $U^{n+1} = 2bU^n$, $U^1 = 6$, is a geometric progression with ratio $2b$ and first term $6$. The general term is $U^n = 6(2b)^{n-1}$. The sequence is convergent for ratio $1 < 2b \leq 1$ and $$\lim_{n \to \infty} U^n = \begin{cases} 0, \quad |2b|<1, \\ 6, \quad 2b = 1. \end{cases}$$