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.
{ (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
Answered by
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
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
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?
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.
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.
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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