radical 2nd n+radical 3th n +...+radical n th All division to n
2006-11-17 07:28:54 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
i am Totally sure that your question is not about a formula that can give the sum for the giving n,coz that formula just doesn't exist!
but if u mean : what happens when n --> ∞ then:
lim ( n^(1/2) + n^(1/3) + .... + n^(1n) ) / n
n -> ∞
= lim (n^(1/2)) / n
n -> ∞
= lim 1 / n^(1/2) = 0
n -> ∞
it means that,as much as u take n bigger,the sum gets much more near to 0.
hope it helps!
2006-11-17 07:58:34 · answer #1 · answered by farbod f 2 · 0⤊ 1⤋
Not totally clear on the end of your statement.
Your looking at a sum:
S=n^(1/2)+n^(1/3)+ n^(1/4) +...., right?
Try factoring out an n:
S = n[n^(-1/2)+n^(-2/3)+n^(-3/4)+...] not much help
I'm not sure about the following either!
exploit the properies of exponents:
exp(S) = exp(n^(1/2))*exp(n^(1/3))*...
= exp(n)^(1/2)*exp(n)^(1/3)*...
=exp(n){exp(n)^(-1/2)
2006-11-17 07:41:45 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
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2016-12-30 14:18:39 · answer #3 · answered by crunkleton 3 · 0⤊ 0⤋
? i dont think many people can help you
2006-11-17 07:35:15 · answer #4 · answered by big.fishes2 2 · 0⤊ 1⤋