Question

Anyone here who is studying Real Analysis, can help me in a sequence exercise? Thank you

Answer 1

sup

Answer 2

What is your question?

Answer 3

Hey @abb0t @zzr0ck3r , what is the upper and lower bound of these finite sum : {1/radical n} , {1/n^3} and {1/(1+n^3)}

Answer 4

I think they all have 1 as an upper bound and 0 as a lower bound, is that correct?

Answer 5

\[\frac{1}{\sqrt{n}}\]?

Answer 6

yes

Answer 7

then yes the upper bound would be 1, this happens when n=1. and as we let n grow the ratio gets smaller because n>1 then sqrt(n) > 1 for n natural.

Answer 8

im with you on \[\frac{1}{n^3}\]is agin 1 when n = 1 and will get smaller as n grows

Answer 9

but
\[\frac{1}{1+n^3}\]the biggest this will get is 1/2

Answer 10

\[2\le1+n^3\]for all n
\[\implies\frac{1}{1+n^3}\le\frac{1}{2}\]

Answer 11

and yes 0 for all the lower bounds

Answer 12

wow thank you very much, is there any method which I should use to proof or I just need to say this and its fine? Thank you again

Answer 13

did it say

"prove the upper bound" or "find the upper bound'

Answer 14

also , there are infinite upper bounds but 1 is the supremum (the smallest upper bound).

Answer 15

it really depends on the teacher as far as proving these
for the first one
\[0\le1\le n^2\text{ for all n in N}\\0\le\frac{1}{n^2}\le1\text{ because }n>0\]

Answer 16

this shows that an upper bound is 1 and a lower bound is 0
but if you need to show that they are the supremum and infimum then you need to do a little more.

Answer 17

ok thank you :D

Answer 18

@zzr0ck3r oh wait we need the lower and upper bound for the finite sum
1+1 + 1+ √2+1/√3 +···+ 1/√n

Answer 19

so a lower bound is 1 and an upper bound is n

Answer 20

what does this have to do with the last one?

Answer 21

i dont see 1/n^3 or 1/(1+n^3)

Answer 22

its not a sequence its a finite sum

Answer 23

for example 1/1 + 1/2^3 + .... + 1/n^3

Answer 24

ahh you have 3 different finite sums

Answer 25

well then n^2 would be an upper bound because each term is amller than n, and we have n terms, so n*n=n^2

Answer 26

n might work al well, but there are infinite answers so....

Answer 27

thank you