# Questions on a recursive sequence

A sequence is defined recursively by

$$b_1=−1, b_{n+1}=((4n−5)/(2n−3))b_n$$.

i)State if it is monotonic

ii)State if it is bounded

iii)Find its limit

For i), $$(b_{n+1}/b_n)>=1$$ gives $$(2n-2)/2n-3)>=0$$, which is positive if $$n>=1$$ and $$n>=3/2$$ (so $$n>=3/2$$), or $$n<=1$$ and $$n<=3/2$$ (so $$n<=1$$). Does this prove that the sequence is increasing? If not, can I prove this by examining the sign of $$f(x) = (2x-2)/(2x-3), x>=1$$, by taking the derivative and the limit to infinity = 1? The question is only worth 2/100 marks, so I guess a quicker answer is expected.

I don't quite have an idea for ii)

For iii), I took $$lim(b_n) = L$$, and then equated it with $$lim(b_{n+1})$$, which gives $$lim(2+(1 /(2n-3)))*L = L$$, therefore $$2L = L$$ and $$L = 0$$. Is this correct?

On a side note, can someone give me a strategy for finding the n-th term? (Here and in general) Thanks!

• For subscripts you need b_{(n+1)} or b_{n+1} not b_(n+1). – David Apr 18 at 0:47
• Thanks, I was actually curious! – JBuck Apr 18 at 0:57

Write the recurrence as $$b_{n+1}=\left(2-\frac 1{2n-3}\right)b_n$$.
Once $$n$$ gets large, you will double every iteration. If you compute the first few terms it is monotonically decreasing.
The limit is $$-\infty$$