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!

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

Doubling every iteration makes sure it is unbounded.

The limit is $-\infty$


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