How can I tell if the sequence\[\left(1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},...\right)\]is compact or not?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

I noticed that the sequence converges to \(1\), which is in the set. I also noticed that every subsequence will also converge to \(1\). To me, it seems the sequence is compact.

- anonymous

yes i think it is compact, which i believe means any subsequence
\[\{x_n\}_{n\geq 1}\] converges to something in the set

- anonymous

of course it is always possible that your subsequence only has a finite number of distinct elements, but that is ok because if such is the case then at least one must occur infinitely often and you can define the subsequence
\[\{x_{n_k}\}=x\] for such an x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

if i recall correcty ( it has been a while) sequence is compact if every subsequence contains a further subsequence that is convergent

- anonymous

That makes sense. :) Thank you

- anonymous

yw
but you still have to say something. for example if you have a subsequence
\[\{x_n\}\] you have to explain why it would have a further subsequence
\[\{x_{n_k}\}\] that converges

- anonymous

so for example you could make
\[\{x_{n_k}\}\] be something like
\[x_k \in \{x_n\} \cap (1-\frac{1}{k},1+\frac{1}{k})\]man that took a long time to type

Looking for something else?

Not the answer you are looking for? Search for more explanations.