Is their a formula to compute a finte harmonic series? 
1+1/2+1/3+1/4...+1/n
For example, if n=75, we can calculate the sums to be 4.9014.
Could be used to solve this teaser:
You are going to drive 75 miles.
You are going to drive the 1st mile at 1 mph,
The 2nd mile at 2 mph,
The 3rd mile at 3 mph,
on and on, until finally,
the 75th mile at 75mph.
How long will it take you to go the 75 miles?

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Clarification of Question by tc2city-ga on 05 Nov 2006 06:24 PST

Obviously, I know I can brute force compute the value once n is given.
I'm looking for some type of formula where n is a variable.(ex.
arithmetic series = n(n+1)/2)

You can get a good approximation by calculating
ln n + .5772.
The larger n is, the better the approximation.

(ln n is the natural, or base-e, logarithm of n;  .5772 is the
approximate value of gamma, or Euler's Constant, which is the limit,
as n increases without bound, of the sum of the series
1+1/2+1/3+...+1/n-ln n.)

For your example with n = 75, the calculation evaluates to about
4.895, which is reasonably close to your value of 4.901.

The formula below gives an even better approximation:

sum = ln n + gamma + 1/(2n) - 1/(12n^2) + 1/(120n^4) ... 

The formula above was found at:
http://en.wikipedia.org/wiki/Euler-Mascheroni_constant

Here is gamma to 45 decimal places, from the same webpage:
? ? 0.57721 56649 01532 86060 65120 90082 40243 10421 59335