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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)]
 one year ago
 one year 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)]
 one year ago
 one year 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
 one year 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?
 one year 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
 one year 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
 one year 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.
 one year ago

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