Finding closed form for this series

Question

I found this beautiful from the book Infinite series of Bromwich, the series is as followed: $\dfrac{x}{24}-\dfrac{x^2}{210}+\dfrac{x^3}{720}-\dfrac{x^4}{1716}+\cdots $

I notice that the denominator can be rewritten as:

$\dfrac{x}{2\cdot3\cdot4}-\dfrac{x^2}{5\cdot6\cdot7}+\dfrac{x^3}{8\cdot9\cdot10}-\dfrac{x^4}{11\cdot12\cdot13} +\cdots$

Is there a closed form for this series?

The general formula for this series is

$$\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}x^{n}}{(3n-1)(3n)(3n+1)}.$$

Answer 1

According to Maple, $$ -\frac{1}3\,\ln \left( 1+x \right) + \frac16 \left( -x^{1/3}-{\frac {1}{x^{1/3}}} \right) \ln \left( 1+x^{1/3} \right) + \frac{1}{12}\left( x^{1/3}+{\frac {1}{x^{1/3}}} \right) \ln \left( 1-x^{1/3}+{x}^{2/3} \right) +\frac12+ \frac{\sqrt{3}}{6}\left( -\,x^{1/3}+{\frac {1}{x^{1/3}}} \right) \arctan \left( {\frac {\sqrt {3}x^{1/3}}{-2+x^{1/3}}} \right) $$

Answer 2

You can rewrite it as:

$$\frac{-1}{2}\left( \sum_{n=1}^{\infty}\frac{(-x)^n}{3n-1} - 2\sum_{n=1}^{\infty}\frac{(-x)^n}{3n} + \sum_{n=1}^{\infty}\frac{(-x)^n}{3n+1}\right)$$

This resembles expansion of logarithm (at least the middle term).

Answer 3

I am doing this followed Donald Splutterwit's advice:

We have:

$$\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}x^{n}}{(3n-1)(3n)(3n+1)}.$$

Let $y^3=x$, we have

$$\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}y^{3n}}{(3n-1)(3n)(3n+1)}.$$

Differentiate it once:

$$\sum_{n=1}^{\infty} \dfrac{(n-2)(-1)^{n-1}(3n)y^{3n-1}}{3(3)(3)}=\sum_{n=1}^{\infty} \dfrac{(n-1)(-1)^{n-1}(3n)y^{3n-1}}{27}$$

I am not sure if this correct, if I differentiate more, the denominator will turn 0, so is it right, or wrong?

Answer 4

Let $x=t^3$, and

$$t\,S(t^3)=\dfrac{t^4}{2\cdot3\cdot4}-\dfrac{t^7}{5\cdot6\cdot7}+\dfrac{t^{10}}{8\cdot9\cdot10}-\dfrac{t^{13}}{11\cdot12\cdot13} +\cdots$$

Now differentiating thrice on $t$,

$$\dddot{t\,S(t^3)} \\=\ddot{S(t^3)+3t^3S'(t^3)} \\=\dot{12t^2S'(t^3)+9t^5S''(t^3)} \\=24tS'(t^3)+81t^4S''(t^3)+27t^7S'''(t^3) \\=t-t^4+t^7-t^{10}+\cdots=\frac t{1+t^3}.$$

(The dots - not quite visible - denote differentiation on $t$; the quotes, differentiation on the argument of $S$).

Hence it "suffices" to solve the ODE

$$24S'(t^3)+81t^3S''(t^3)+27t^6S'''(t^3)=\frac1{1+t^3}$$

or

$$24S'(x)+81xS''(x)+27x^2S'''(x)=\frac1{1+x}.$$

It is of the Euler-Cauchy type.