Surprisingly long closed form for simple series

Question

For natural $A$ define $$ f(A)=\sum_{n=1}^\infty \frac{1}{A^n}\left(\frac{1}{An+1}- \frac{1}{An+A-1}\right)$$

$f(A)$ is BBP (Bailey-Borwein-Plouffe) formula and allows digit extraction in base $A$.

For small values of $A$, Wolfram Alpha finds closed form in simple terms of 2F1 and also very long closed form in terms of algebraic numbers and logarithms.

Link to f(5) and link to f(7)

The elementary closed for for $f(5)$ is about one screen long, here it is in plaintext:

-3/4 - log(1 - 1/5^(1/5))/5^(4/5) - ((-1)^(2/5) log(1 - e^(-(2 i π)/5)/5^(1/5)))/5^(4/5) + ((-1)^(3/5) (log(-1 + e^((2 i π)/5)/5^(1/5)) - i π))/5^(4/5) - (-1/5)^(4/5) log(1 - e^(-(4 i π)/5)/5^(1/5)) + 
((-1)^(1/5) (log(-1 + e^((4 i π)/5)/5^(1/5)) - i π))/5^(4/5) + 1/4 ((4 log(1 - 1/5^(1/5)))/5^(1/5) - (4 (-1)^(3/5) log(1 - e^(-(2 i π)/5)/5^(1/5)))/5^(1/5) + (4 (-1)^(2/5) (log(-1 + e^((2 i π)/5)/5^(1/5)) - i π))/5^(1/5) 
- 4 (-1/5)^(1/5) log(1 - e^(-(4 i π)/5)/5^(1/5)) + (4 (-1)^(4/5) (log(-1 + e^((4 i π)/5)/5^(1/5)) - i π))/5^(1/5))

The elementary closed form for $f(7)$ is several screens long.

Q1 What is the intuition for so long elementary closed form?

Q2 Is there simpler elementary closed form for $f(A)$ in terms of $A$?

Answer 1

Let $a:=A$. For $a\ge2$ and $r\in\{0,\dots,a-1\}$, we have $$s(r):=\sum_{n=1}^\infty\frac1{a^n}\frac1{an+r} =\sum_{n=1}^\infty\frac1{a^n}\int_0^1 dx\, x^{an+r-1} \\ =\int_0^1 dx\, \sum_{n=1}^\infty\frac1{a^n}x^{an+r-1} =\int_0^1 dx\, R(x),$$ where $$R(x):=\frac{x^{a+r-1}/a}{1-x^a/a}.$$

Using the partial fraction decomposition of the rational function $R$, we get an elementary expression (of length $\asymp a$) of the sum $s(r)$.

The sum in question is then $s(1)-s(a-1)$, for $a\ge2$. The case $a=1$ is easier.

Answer 2

The intuition becomes clear when we evaluate the series by bare hands.

For $r\in\{1,\dotsc,A\}$ we have \begin{align*}\sum_{n=0}^\infty \frac{1}{A^n(An+r)}&= \sum_{\substack{m\geq 1\\m\equiv r\pmod{A}}}\frac{1}{A^{(m-r)/A}m}\\ &=\sum_{m=1}^\infty\frac{1}{A^{(m-r)/A}m}\left(\frac{1}{A}\sum_{k=1}^A e^{2\pi i (m-r)k/A}\right)\\ &=\frac{1}{A^{1-r/A}}\sum_{k=1}^Ae^{-2\pi i kr/A}\sum_{m=1}^\infty\frac{e^{2\pi ikm/A}}{A^{m/A}m}.\end{align*} The inner sum on the right-hand side equals $-\log(1-e^{2\pi ik/A} A^{-1/A})$, and dropping the $n=0$ term is easy.

Answer 3

$$f(A)=\frac 1{A^2}\left(\Phi \left(\frac{1}{A},1,1+\frac{1}{A}\right)-\Phi \left(\frac{1}{A},1,2-\frac{1}{A}\right)\right)$$ where $\Phi$ is Lerch function.