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Limiting Mean (Posted on 2010-03-07) Difficulty: 2 of 5

Limit H(m)/m
m → ∞

where, H(m) denotes the harmonic mean of the m positive integers m+1, m+2, ....., 2m.

As a bonus, evaluate this limit:

Limit R(m)/m
m → ∞

where, R(m) denotes the root mean square (RMS) of the m positive integers m+1, m+2, ....., 2m.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts Partial (hon-rigorous) solution? (ho bonus) | Comment 1 of 3

The harmonic mean for the values in this problem is equal to:

m/[ (x=m+1 to 2m) 1/x.  Divide this by m and you get

1/ (x=m+1 to 2m) 1/x. 

If you take the limit, this starts to look a lot like a definite integral. 

1 / {integral} m+1 to 2m of 1/x dx.

If you grant me this, then you get:

1/{ln(2m) - ln(m)} = 1/ln(2).

A spreadsheet check for m=1000 amd 10000 suggests strongly that this is the answer that is converged upon.

Sorry - my math is not any better than this.  I trust that someone can be more rigorous than I.

Answer:  1/(ln(2)





  Posted by Kenny M on 2010-03-07 22:30:59
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