Posted by Paul on Thursday, September 20, 2012 at 10:13am.
(n+1)^2/(n^2+1) = 1 + 2n/(n^2+1)
the sequence obviously increases, and the limit is 1
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?
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
so the upper bound is 1, since it converges, right?
let's see. every term is greater than 1. Why would you say 1 is the upper bound?
1 is the lower bound, no?
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.
Oh, 2 would be the upper bound? Because it can't go higher than 2?
or rather it starts off at 2
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.
Ok so lower bound is the lowest achievable value, so upper bound will be the highest, right?
2 is an upper bound, but it's boring. we are interested in what happens at the tail end of the sequence.
Okay I get it now!! Thanks a lot 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.
so what's the best way to find that number? would it be putting n towards infinity or 0?