Asked by Paul

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.
Answered by Steve
(n+1)^2/(n^2+1) = 1 + 2n/(n^2+1)

the sequence obviously increases, and the limit is 1
Answered by 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?
Answered by 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
Answered by Paul
so the upper bound is 1, since it converges, right?
Answered by 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?
Answered by 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.
Answered by Paul
Oh, 2 would be the upper bound? Because it can't go higher than 2?
Answered by Paul
or rather it starts off at 2
Answered by 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.
Answered by Paul
Ok so lower bound is the lowest achievable value, so upper bound will be the highest, right?
Answered by Steve
2 is an upper bound, but it's boring. we are interested in what happens at the tail end of the sequence.
Answered by Paul
Okay I get it now!! Thanks a lot Steve :)
Answered by 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.
Answered by Paul
so what's the best way to find that number? would it be putting n towards infinity or 0?
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