Question
Determine if the following sequence:
{ (n + 1)^2 / (n^2 + 1) },
is ascending, descending and find the lower bound b OR the upper bound B.
(n + 1)^2 / (n^2 + 1)
>
((n+1) + 1)^2 / ((n+1)^2 + 1)
so the sequence is descending,
it's the bounds I don't know how to find.
{ (n + 1)^2 / (n^2 + 1) },
is ascending, descending and find the lower bound b OR the upper bound B.
(n + 1)^2 / (n^2 + 1)
>
((n+1) + 1)^2 / ((n+1)^2 + 1)
so the sequence is descending,
it's the bounds I don't know how to find.
Answers
Steve
(n+1)^2/(n^2+1) = 1 + 2n/(n^2+1)
the sequence obviously increases, and the limit is 1
the sequence obviously increases, and the limit is 1
Paul
but if I put n = 1 in both I get =2 for the first and =1.8 for n+1. Or is that not how I go about doing it?
Steve
Oops. The sequence decreases, since 2n < n^2+1
The sequence goes 2, 1.8, 1.6, 1.47, 1.39, 1.32 ... converging to 1
The sequence goes 2, 1.8, 1.6, 1.47, 1.39, 1.32 ... converging to 1
Paul
so the upper bound is 1, since it converges, right?
Steve
let's see. every term is greater than 1. Why would you say 1 is the upper bound?
1 is the lower bound, no?
1 is the lower bound, no?
Paul
Because I thought that if it decreases it'll eventually get lower than 1. What is the definition of lower and upper bound, I don't get that part.
Paul
Oh, 2 would be the upper bound? Because it can't go higher than 2?
Paul
or rather it starts off at 2
Steve
note that each term is 1 + a fraction.
as the fraction decreases, the terms approach 1, but are always slightly greater than 1.
Thus, the minimum value achievable is 1, which makes 1 a lower bound. In fact, it is the greatest lower bound: there is no number greater than 1 which is less than all terms of the sequence.
as the fraction decreases, the terms approach 1, but are always slightly greater than 1.
Thus, the minimum value achievable is 1, which makes 1 a lower bound. In fact, it is the greatest lower bound: there is no number greater than 1 which is less than all terms of the sequence.
Paul
Ok so lower bound is the lowest achievable value, so upper bound will be the highest, right?
Steve
2 is an upper bound, but it's boring. we are interested in what happens at the tail end of the sequence.
Paul
Okay I get it now!! Thanks a lot Steve :)
Steve
the lower bound is 1, but that is not the lowest achievable value. A lower bound is a number which is less than or equal to any term of the sequence.
But this sequence never has a term which is 1. All terms are greater than 1, but there is always a number closer to 1 than any amount you specify.
Thus 1 is a lower bound, but it is not a term of the sequence. 2n/(n^2+1) is never zero.
But this sequence never has a term which is 1. All terms are greater than 1, but there is always a number closer to 1 than any amount you specify.
Thus 1 is a lower bound, but it is not a term of the sequence. 2n/(n^2+1) is never zero.
Paul
so what's the best way to find that number? would it be putting n towards infinity or 0?