Generalisation of $\log(1-x)$ expansion?

Question

The standard logarithm expansion is given by

\begin{equation} \log(1-x) = -\sum_{n=1}^{\infty}\frac{x^n}{n}, \qquad |x|<1 \end{equation}

I have a sum which is of the form

\begin{equation} \sum_{n=1}^{\infty}\frac{x^n}{n+\frac{1}{m}}, \end{equation}

where $m$ is some integer. Is there some known function which has this kind of expansion? Given that standard Taylor series expansions always have $1/n!$ factors, it's quite hard to see how something like these fractional coefficients could appear.

Answer 1

The best I can think of is start with: $$\sum_{n=1} ^{\infty} x^n = \frac{1}{1-x}-1,$$ then multiplying both sides by $x^{(\frac{1}{m}-1)}$ to get $$\sum_{n=1} ^{\infty} x^{(n+\frac{1}{m}-1)} = \frac{x^{(\frac{1}{m}-1)}}{1-x} - x^{(\frac{1}{m}-1)}.$$ After integrating from $0$ to $x$ we have: $$\sum_{n=1} ^{\infty} \frac {x^{(n+\frac{1}{m})}}{n+\frac{1}{m}} = \int_0 ^x \frac{t^{(\frac{1}{m}-1)}}{1-t} \cdot dt - m \cdot x^{\frac{1}{m}},$$ which can be divided by $x^{\frac{1}{m}}$ to become: $$\sum_{n=1} ^{\infty} \frac {x^n}{n+\frac{1}{m}} = x^{-\frac{1}{m}} \cdot \int_0 ^x \frac{t^{(\frac{1}{m}-1)}}{1-t} \cdot dt - m$$ Not very pretty, but I doubt that you can find a better solution without the usage of special functions. Well, I hope it helps you anyway.

Answer 2

Here is a silly way of doing it. Start with

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\;\;\;\;|x|<1$$

Replace $x\mapsto x^m$ and get

$$\frac{1}{1-x^m}=\sum_{n=0}^\infty x^{mn},\;\;\;\;|x|<1$$

Integrate from $0$ to $x$ and get

$$\int_0^x\frac{1}{1-t^m}dt=\sum_{n=0}^\infty\frac{x^{mn+1}}{mn+1},\;\;\;\;|x|<1$$

Substitute $x\to x^{1/m}$ and multiply both sides by $m$ to get

$$m\int_0^{x^{1/m}}\frac{1}{1-t^m}dt=\sum_{n=0}^\infty\frac{x^{n+1/m}}{n+\frac{1}{m}}=mx^{1/m}+x^{1/m}\sum_{n=1}^\infty\frac{x^n}{n+\frac{1}{m}},\;\;\;\;|x|<1$$

As such,

$$\sum_{n=1}^\infty\frac{x^n}{n+\frac{1}{m}}=m\frac{\int_0^{x^{1/m}}\frac{1}{1-t^m}dt-x^{1/m}}{x^{1/m}},\;\;\;\;|x|<1$$

Solutions now completely depend on $m$. To check, consider $m=1$ and get

$$\sum_{n=1}^\infty\frac{x^n}{n+1}=\frac{\int_0^x\frac{1}{1-t}dt-x}{x}=\frac{\ln(1-x)-x}{x}$$

For $m=2$,

$$\sum_{n=1}^\infty\frac{x^n}{n+\frac{1}{2}}=2\frac{\int_0^{\sqrt{x}}\frac{1}{1-t^2}dt-\sqrt{x}}{\sqrt{x}}=\frac{2}{\sqrt{x}}\left(\frac{1}{2}\ln\left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)-\sqrt{x}\right)$$

There is not much else we can do in a reasonable amount of time, but the key is that all these denominators are factorable (even if we have to use complex numbers) and so we can find the solution in terms of a finite sum of complex numbers, which is at least a little better.