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October 31, 2014

October 31, 2014

Posted by **Paul** on Thursday, September 20, 2012 at 10:13am.

{ (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**, Thursday, September 20, 2012 at 12:23pm(n+1)^2/(n^2+1) = 1 + 2n/(n^2+1)

the sequence obviously increases, and the limit is 1

- Calculus - Series4 -
**Paul**, Thursday, September 20, 2012 at 12:31pmbut 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**, Thursday, September 20, 2012 at 1:06pmOops. 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**, Thursday, September 20, 2012 at 1:09pmso the upper bound is 1, since it converges, right?

- Calculus - Series4 -
**Steve**, Thursday, September 20, 2012 at 2:05pmlet'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**, Thursday, September 20, 2012 at 2:11pmBecause 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**, Thursday, September 20, 2012 at 2:12pmOh, 2 would be the upper bound? Because it can't go higher than 2?

- Calculus - Series4 -
**Paul**, Thursday, September 20, 2012 at 2:12pmor rather it starts off at 2

- Calculus - Series4 -
**Steve**, Thursday, September 20, 2012 at 2:18pmnote 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**, Thursday, September 20, 2012 at 2:20pmOk so lower bound is the lowest achievable value, so upper bound will be the highest, right?

- Calculus - Series4 -
**Steve**, Thursday, September 20, 2012 at 2:20pm2 is an upper bound, but it's boring. we are interested in what happens at the tail end of the sequence.

- Calculus - Series4 -
**Paul**, Thursday, September 20, 2012 at 2:21pmOkay I get it now!! Thanks a lot Steve :)

- Calculus - Series4 -
**Steve**, Thursday, September 20, 2012 at 2:23pmthe 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**, Thursday, September 20, 2012 at 2:26pmso what's the best way to find that number? would it be putting n towards infinity or 0?

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