$\{a_n\}$ and $\{b_n\}$ are two series of real number with $a_0=2$,$a_1=3$ and $a_{n+1}=3a_n-a_{n-1},\forall n\in \mathbb{N^+}$,when $m= 2^n$, we have $\sqrt{b_n}=\frac{m}{a_m}$.Now, for the series $$\sum{b_n}$$ is there a real number $\tau$, such that $$\sum{b_n}=\tau$$ and find the exact number? Here is my work: By add $\frac{3-\sqrt{5}}{2}a_n$ to the equation:$$a_{n+1}=3a_n-a_{n-1}$$ I hereby make sure that $a_{n+1}=\frac{3-\sqrt{5}}{2}a_n+q^n\sqrt{5}$,with $q=\frac{3+\sqrt{5}}{2}$. it seems that by matching the coefficients again, I figure out the general term formula for $a_n$, but the general term formula seems to be very complicated. I wonder if there's an easy way to get around figuring out the general formula?
Update: I have figured out that $a_n=(\frac{3-\sqrt{5}}{2})^n+(\frac{3+\sqrt{5}}{2})^n$ but i cant push any furthur.
You've got $a_n=q^n+q^{-n}\color{gray}{\text{ with }q=(3+\sqrt5)/2}$, which implies $a_{2n}=a_n^2-2$. If we put $c_n=a_{2^n}$, we then get $c_0=3$ and $c_{n+1}=c_n^2-2$; note that $b_n=4^n/c_n^2$.
Now we claim for $n\geqslant 0$ that $$\sum_{k=0}^{n-1}b_k=\frac15-\frac{4^n}{c_{n+1}-2}$$ (assuming that the LHS is zero at $n=0$). Indeed, for $n>0$, by induction, $$\sum_{k=0}^{n-1}b_k=\frac15-\frac{4^{n-1}}{c_n-2}+b_{n-1}=\frac15-\frac{4^{n-1}}{c_n-2}+\frac{4^{n-1}}{c_n+2}$$ gives the expected expression. After taking $n\to\infty$, we obtain finally $\color{blue}{\sum_{n=0}^\infty b_n=1/5}$.