[tex]\displaystyle\dfrac{5^n+1}{9^n}=\left(\dfrac59\right)^n+\dfrac1{9^n}[/tex]
As [tex]n\to\infty[/tex], both terms will approach 0, so the sequence converges. To show this, we can invoke the monotone convergence theorem.
Consider the function [tex]f(a,x)=a^x[/tex], where [tex]0<a<1[/tex]. We have
[tex]\dfrac{\partial f(a,x)}{\partial x}=\ln a\,a^x=\ln a\,f(a,x)[/tex]
and since [tex]f(a,x)>0[/tex] for any such [tex]a[/tex], and [tex]\ln a<0[/tex], we have that [tex]\dfrac{\partial f(a,x)}{\partial x}<0[/tex], which means the function is decreasing over its entire domain.
To recap: [tex]f(x)>0[/tex] for all [tex]x[/tex], so both [tex]f\left(\dfrac59,x\right)[/tex] and [tex]f\left(\dfrac19,x\right)[/tex] are bounded below. Then [tex]\dfrac{\partial f(a,x)}{\partial x}<0[/tex] for all [tex]0<a<1[/tex], which means [tex]f\left(\dfrac59,x\right)[/tex] and [tex]\f\left(\dfrac19,x\right)[/tex] are decreasing.
Therefore by the monotone convergence theorem, both sequences converge to 0, and so must [tex]a_n[/tex].
