I have x_n =[1/sqrt(2)+1/sqrt(3)+....+1/sqrt(n)]/sqrt(n) I must find lim of a_n with stoltz-cesaro so [1/sqrt(2)+1/sqrt(3)+....+1/sqrt(n)]=a_n and sqrt(n) = b_n and in b_n -> infinity i must show that lim a_n/b_n has the same limit as [a_(n+1) -a_(n)]/[b_(n+1) - b_(n)]

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.

A community for students.

I have x_n =[1/sqrt(2)+1/sqrt(3)+....+1/sqrt(n)]/sqrt(n) I must find lim of a_n with stoltz-cesaro so [1/sqrt(2)+1/sqrt(3)+....+1/sqrt(n)]=a_n and sqrt(n) = b_n and in b_n -> infinity i must show that lim a_n/b_n has the same limit as [a_(n+1) -a_(n)]/[b_(n+1) - b_(n)]

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

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

are you asked to prove stoltz-cesaro or just apply it here? if it's the latter then just take your limit: \[\lim_{n \rightarrow \infty} \frac{ a_{n+1} - a_n }{ b_{n+1} - b_n } = \lim_{n \rightarrow \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}\] then just rationalize the denominator and you should get a pretty simple answer
it's good i used your lim and i got \[1 + \frac{ \sqrt{n} }{ \sqrt{n+1} }\] what can i do next this is not looking like a final form and i have a hint that the answer is two, can this be true?
ugh take the limit? and the 2nd term is 1 when u do that

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

is not infinity over infinity? i am just asking i don' t know
ok when n is going to infinity, that 1 in n+1 becomes negligible, so pretty much u have sqrt(n)/sqrt(n), which is just 1.
i understood, thank you very much :D

Not the answer you are looking for?

Search for more explanations.

Ask your own question