Finding a good comparison/bound for determining the convergence of a series The series is defined as follows: $b_0=1,b_1=-7,b_k=2b_{k-1}+b_{k-2}$. I need to find a good comparison sequence to determine whether $\sum_{k\geq0}1/b_k$ converges. I considered using $1/k^2$, which intuitively $1/k^2\gg1/b_k$ for all $k>0$, but the proof that this is the case is harder than I expected. Definitely possible with multiple proofs by induction which would take up more than one page. Or is there a simpler proof I'm missing?
I would definitely appreciate hints as to a good comparison series. And I'd also like to know what kind of thought processes suggested that such a series would be a good comparison.
 A: Some thoughts: I am not fond of negative numbers, so let us keep the recurrence and let $a_0=-1$ and $a_1=7$. That just switches all the signs. 
It may now be a good idea to write down the first few terms. Because of the $2a_{k-1}$ part, the terms (except at the beginning) keep more than doubling, so we can compare the $\frac{1}{a_k}$ with $\frac{1}{2^k}$. 
Remark: The $\frac{1}{a_k}$ decrease faster than $\frac{1}{2^k}$, and we can in fact get very precise information about the rate. However, if we are just interested in the issue of convergence, an easy to find bound that does the job is good enough.
A: You can solve $b_n$ for general form and find a suitble sequnece to compare whor it.
$$x^k=2x^{k-1}+x^{k-2}$$
So
$$x^2=2x+1$$
And
$$x_1= \frac{2+\sqrt 8}{2} , x_2= \frac{2-\sqrt 8}{2}$$
Then 
$$b_n =(\frac{1}{2}-2\sqrt 2)(1+\sqrt 2)^n +(\frac{1}{2}+2\sqrt 2)(1-\sqrt 2)^n$$
If $a_n= 3^n  (\frac{1}{2}-2\sqrt 2)+1$ then $b_n >a_n$ and since $\sum \frac{1}{a_n}$ is converges , we claim that $\sum \frac{1}{b_n}$ is converges
