Question

\[A\] is a subset of \[\mathbb{R} \quad \Large{A:=\{\frac{1}{n^{2}+4} \quad | \quad n \in \mathbb{N}_{+}\}}\] determine wheter set is bounded from above or below and deterimine if there is sup or inf exists

Answer 1

the set is bounded below

Answer 2

can you prove it pls ?

Answer 3

txh loser but i need more detailed description

Answer 4

I need a prove like by attachment

Answer 5

ok that sounds better, thank you loser ;)

Answer 6

its weird when you say loser

Answer 7

@jonask I don't know, I was taught that, if it's weird to you, so, I take it off .

Answer 8

hey Loser whats problem i need the solution pls add back ;) i think there is some missunderstanding between you and jonask

Answer 9

@tunahan let me tag other for help. I am not good at explanation. @cwrw238

Answer 10

please!!

Answer 11

i was happy with your explanation :)

Answer 12

let wait for hartn. he is good at explanation. I knew it

Answer 13

ok good :)

Answer 14

As diverge test states that lim A_n when n goes to infinitive is not 0, then A_n diverges.
therefore, I take lim A_n = lim (1/n^2 +4) =0 when n goes to infinitive. That includes that A_n converges to 0 , In other words, 0 is greatest low bound of A_n
I don't delete it.

Answer 15

thx

Answer 16

@hartnn think of this, if I have A_n is set of numbers such that A_n = 1/n^2 +4. if I have n=1 , A_1 = 1/5
n=2, A_2 = 1/8
n=3 A_3 = 1/13 so on and so on .... so that my subset A goes to where??
A ={ 1/5,1/8/,1/13.......} it goes closer and closer to 0. Although it never get 0 , but 0 is the point it will come up. think about the function down, down down...... but never negative. therefore, line 0 is the greatest low bound.