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graydarl
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I have x_n =[1/sqrt(2)+1/sqrt(3)+....+1/sqrt(n)]/sqrt(n)
I must find lim of a_n with stoltzcesaro
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)]
 2 years ago
 2 years ago
graydarl Group Title
I have x_n =[1/sqrt(2)+1/sqrt(3)+....+1/sqrt(n)]/sqrt(n) I must find lim of a_n with stoltzcesaro 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)]
 2 years ago
 2 years ago

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cnknd Group TitleBest ResponseYou've already chosen the best response.1
are you asked to prove stoltzcesaro 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
 2 years ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
ugh take the limit? and the 2nd term is 1 when u do that
 2 years ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
is not infinity over infinity? i am just asking i don' t know
 2 years ago

cnknd Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
i understood, thank you very much :D
 2 years ago
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