How do you find the sum of a set of fractions where the denominator is increasing linearly?

Question

Is there a quick way to find the sum of fractions where the denominator increases linearly within a finite range and infinitely?

For example:

1/5 + 1/6 + 1/7 .....

or:

1/5 + 1/7 + 1/9 .... 1/101

Answer 1

There is no quick way.

This is a generalization of a sub segment of the harmonic series, and while that can be estimated asymptotically, it cannot be computed exactly.

Answer 2

You can use an integral representation

$$\sum_{n=0}^{N-1}\frac{1}{An+B}=\int_0^1\frac{x^{AN+B}-x^{B}}{x(x^{A}-1)}dx$$

Though this doesn't really help to compute the values.

Answer 3

You can approximate the result using Euler-MacLaurin summation formula $$\sum_{n=a}^b f(n) = \int_a^b f(x)\,dx +\frac{f(a)+f(b)}{2} + \frac{1}{6}\frac{f'(b) - f'(a)}{2!} +\cdots $$

So, if $$f(n)=\frac 1 {\alpha n +\beta}$$ $$\sum_{n=a}^b f(n) =\frac 1 \alpha \log \left(\frac{\alpha\, b+\beta }{ \alpha \,a+\beta }\right)+ \left(\frac{1}{ \alpha\,a +\beta }+\frac{1}{\alpha \, b+\beta }\right)+\frac \alpha{12}\left(\frac{1}{(a \alpha +\beta )^2}-\frac{1}{(\alpha b+\beta )^2}\right)+\cdots$$

For your second case where $\alpha=1$, $\beta=0$, $a=5$, $b=101$, the approximation will be $$\log \left(\frac{101}{5}\right)+\frac{53}{505}+\frac{848}{255025}+\cdots=\color{red}{3.1139}58$$ to be compared to the exact value $$\frac{877105921057330616800795116172051213424683}{281670315928038407744716588098661706369472}=\color{red}{3.113945}$$ which corresponds to a relative error of $0.0004$%.

And we can do better adding the next derivatives.