Guest Posted December 30, 2009 Report Share Posted December 30, 2009 Evaluate this limit: Limit F(m)/m m → ∞ where, F(m) denotes the root mean square (RMS) of the m positive integers m+1, m+2, ....., 2m. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 30, 2009 Report Share Posted December 30, 2009 If you take the square of the limit, then you car write the expression in the form (1/m)*( [(m+1)/m]^2+...+[(2m)/m]^2 ). When m goes to infinity, this is the integral of x^2 from 1 to 2. So the final answer should be sqrt(7/3) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 30, 2009 Report Share Posted December 30, 2009 I found a formula for rms(x to 2x)/x it is sqrt((14x^3+15x^2+x)/(6x))/x as x approaches infinity, the limit is 1.527525313... or sqrt(7/3) Quote Link to comment Share on other sites More sharing options...
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Evaluate this limit:
Limit F(m)/m
m → ∞
where, F(m) denotes the root mean square (RMS) of the m positive integers m+1, m+2, ....., 2m.
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