Posted by Paul on .
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.

Calculus  Series4 
Steve,
(n+1)^2/(n^2+1) = 1 + 2n/(n^2+1)
the sequence obviously increases, and the limit is 1 
Calculus  Series4 
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?

Calculus  Series4 
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 
Calculus  Series4 
Paul,
so the upper bound is 1, since it converges, right?

Calculus  Series4 
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? 
Calculus  Series4 
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.

Calculus  Series4 
Paul,
Oh, 2 would be the upper bound? Because it can't go higher than 2?

Calculus  Series4 
Paul,
or rather it starts off at 2

Calculus  Series4 
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. 
Calculus  Series4 
Paul,
Ok so lower bound is the lowest achievable value, so upper bound will be the highest, right?

Calculus  Series4 
Steve,
2 is an upper bound, but it's boring. we are interested in what happens at the tail end of the sequence.

Calculus  Series4 
Paul,
Okay I get it now!! Thanks a lot Steve :)

Calculus  Series4 
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. 
Calculus  Series4 
Paul,
so what's the best way to find that number? would it be putting n towards infinity or 0?