Question

Given \(a_1 = 1\), \(b_1=-\frac{1}{2}\), and
\(a_n = \frac{a_{n-1}}{b_{n-1}}\) and \(b_n = a_{n-1}*b_{n-1}\) for all n>1.
Show that \((a_n)_{n\in\mathbb{N}}\) and \((b_n)_{n\in\mathbb{N}}\) diverge.

Answer 1

\[ a_n\cdot b_n = \frac{a_{n-1}}{b_{n-1}}\cdot a_{n-1}b_{n-1} = (a_{n-1})^2 \]

similarly,
\[ \frac{a_n}{ b_n} = \frac{a_{n-1}}{b_{n-1}}\cdot \frac 1{ a_{n-1}b_{n-1}} = \frac 1 {b_{n-1}^2}\]

Answer 2

\[ a_n = \frac{a_{n-1}}{b_{n-1}} = \frac{1}{b_{n-1}^2}\]
and
\[ b_n =a_{n-1}b_{n-1} = a_{n-2}^2 \]

Answer 3

*
\[
a_n = \frac{a_{n-1}}{b_{n-1}} = \frac{1}{b_{n-2}^2}
\]

Answer 4

\[ a_{n} = \frac{1}{(a_{n-4})^4}\]

Answer 5

\[ a_2 = -2, a_6 = 1/16, a_{10} = 16^4 \dots \]

Answer 6

\[ a_{18} = (-2)^{4^3} \\
\huge {a_{8n+2} = (-2)^{4^{(2n)}}}\]

easy to see that it diverges as n->infty

Answer 7

similarly it can be shown via recursion relation for b_n
GTG see you later

Answer 8

Hmm.... So, I can show the subsequence of it diverges and claim that it diverges by the comparison test!!!

Answer 9

yep

Answer 10

a sequence converges if and only if all it's sub-sequence converges ... or you can use the cauchy sequence test too.

Answer 11

if particular if \( a_{f(n)} \) is sub-sequence of \( a_n \) then \( a_n \) converges if \( a_{f(n)} \) converges for all \(f:\Bbb N \to \Bbb N \).

Answer 12

*iff