Sequence satisfying a condition for all n

1. Sep 22, 2009

StarTiger

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

Suppose that a sequence {s_n} of positive numbers satisfies the condition s_(n+1) > αs_n for all n where α > 1. Show that s_n → ∞

My teacher mentioned something about making it into a geometric sequence and taking the log. I'm just confused.

2. Relevant equations

3. The attempt at a solution

2. Sep 22, 2009

LCKurtz

You can start with s2 > as1. Now what about s3? Can you compare it to s2 and s1? Continue...

3. Sep 22, 2009

HallsofIvy

Staff Emeritus
So $s_2> a s_n$, $s_3> a s_2> a(a s_1)= a^2 s_1$, $s_4> a s_3> a(a^2 s_1)= a^3 s_1$. So $s_n>$ a to what power times $s_1$? What does that have to do with a "geometric sequence"?

4. Sep 22, 2009

fmam3

If $$(s_n)$$ is a sequence and the limit $$\lim_{n \to \infty}|s_{n+1} / {s_n}| = L$$ exists and $$L < 1$$, then $$\lim s_n$$ converges. If not, what do you think happens?