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graydarl

  • 3 years ago

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)]

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  1. cnknd
    • 3 years ago
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    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

  2. graydarl
    • 3 years ago
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    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?

  3. cnknd
    • 3 years ago
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    ugh take the limit? and the 2nd term is 1 when u do that

  4. graydarl
    • 3 years ago
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    is not infinity over infinity? i am just asking i don' t know

  5. cnknd
    • 3 years ago
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    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.

  6. graydarl
    • 3 years ago
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    i understood, thank you very much :D

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