Summing Finitely Many Terms of Harmonic Series: $\sum_{k=a}^{b} \frac{1}{k}$ How do I calculate sum of a finite harmonic series of the following form?
$$\sum_{k=a}^{b} \frac{1}{k} = \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \cdots + \frac{1}{b}$$
Is there a general formula for this? How can we approach this if not?

The formula for sum of H.P. remained unknown for many years, but now I have found the formula as an infinite polynomial.
The formula or the infinite polynomial which is
equal to the sum of th H.P. $\frac{1}{a} + \frac{1}{a+d} + \frac{1}{a+2d} + \frac{1}{a+3d} + ........$ is :-
$Sum of H.P. = (1/a)  [1 + \frac{1}{1×(1+b)}(x-1) - \frac{b}{(2×(1+b)(2+b)}(x-1)(x-2) + \frac{b^2}{3×(1+b)(2+b)(3+b)}(x-1)(x-2)(x-3) - \frac{b^3}{4×(1+b)(2+b)(3+b)(4+b)}(x-1)(x-2)(x-3)(x-4) + \frac{b^4}{5×(1+b)(2+b)(3+b)(4+b)(5+b)}(x-1)(x-2)(x-3)(x-4)(x-5) - .....]$
Here, $b=d/a$ and $x$ is the number of terms upto
which you want to find the sum of the H.P.
Substitute any natural number in place of x and
get the sum of the harmonic series.
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I have not provided any proof here since that will be a very difficult task at this platform. 
