Modification to geometric infinite summation

I've been exploring the collatz conjecture ($3n+1$ problem) as an interest and came up with the following problem. Would greatly appreciate if help could be found!

Typically, for geometric summations to infinity, typical textbooks deal with questions in the form of: $$x = \sum_{i=1}^{\infty}\frac{1}{r^{i}}.$$

I'd like to know what would happen if, instead, the summation was in the form of: $$x = \sum_{i=1}^{\infty}\frac{1}{r^{i}+k},$$ e.g. $$\frac{1}{3}+\frac{1}{15}+\frac{1}{63}+...$$ for the case of $r=4, k=-1$.

In this case, would it be easy/possible to compute the infinite sum? Unfortunately, I have not even been able to find a name for this kind of problem. Also, given that the infinite sum for $r^i$ converges for positive $r$, this series ought to converge for positive $k$ as well. Whether or not it would converge for negative $k$, however, is another (though also interesting) question.

• $r=10,k=1$ wouldn't give the series you wrote. – Deepak Aug 10 '17 at 10:29
• @Deepak doesn't it? $1/(10^0+1)+1/(10^1+1)+1/(10^2+1)+...$ – Ducky Aug 10 '17 at 11:14
• @Ducky OP changed the question, see the edit history. – Deepak Aug 10 '17 at 11:17
• Mathematica gives $$\sum_{i=0}^{\infty}\frac{1}{r^i+k}=\frac{2 \psi _r\left(-\frac{\log (-k)}{\log (r)}\right)+2 \log (-k)+2 \log (r-1)+\log(r)}{2 k \log (r)}$$ where $\psi _r$ is the q-digamma function. Not sure how to derive such a result though. – James Arathoon Aug 10 '17 at 12:08
• Oh, right, it isn't $\frac{1}{1}+\frac{1}{11}+\frac{1}{111}$... Sorry, was thinking of something else - I think something along the lines of $r=10,k=-1$, and then multiply the whole thing by $\frac{1}{9}$. Also renders the point about convergence for positive k moot then - but from James' comment, positive k would be uncomputable due to the log. Will edit the post accordingly then. – brorion Aug 10 '17 at 15:14