How can I represent the summation $\frac{x}{1^2}+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\cdots$

Question

I'm looking to represent the summation $\frac{x}{1^2}+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\cdots$ in the form of

$$\frac{3}{2}\arctan^2\left(\sqrt3(a_1x+a_2x^2+ a_3x^3+ \cdots)\right)$$

I wanted to represent the Basel problem as an inverse trigonometric function. (Here at x=1, the value $(a_1+a_2+ a_3+ \cdots)$ must equal 1, so both sides equal $\pi^2/6$)

If the $,a_1,a_2,a_3...$ coefficients followed a set pattern, the Basel problem may alternatively be proved by proving that the sum of coefficients equals $1$.

I attempted to evaluate the coefficients by comparing derivative values of LHS and RHS at x=$0$, however the complexity soon skyrocketed.

Is there a more efficient way of evaluating the Taylor series, like substituting x as another trigonometric function or is the problem inherently not representable using elementary functions?

Answer 1

You must take care that $$\frac{3}{2}\arctan^2\left(\sqrt3\,\sum _{i=1}^3 a_i x^i\right)-\text{Li}_2(x)$$ write $$-x+\left(\frac{9}{2} a_1^2-\frac{1}{4}\right) x^2+\left(9 a_1 a_2-\frac{1}{9}\right) x^3+O\left(x^4\right)$$

Choosing $a_1=\frac{1}{3 \sqrt{2}}$, the other $a_i=\frac {b_i}{3\sqrt 2}$, the very first $b_i$ are $$\left\{1,\frac{2}{9},\frac{101}{648},\frac{6907}{72900},\frac{1393 087}{20995200},\frac{8966875}{185177664},\frac{204870041231}{55 55329920000},\cdots\right\}$$

Answer 2

The Sum cannot be represented in terms of elementary functions, however the Polylogarithm $\operatorname{Li}_s(z)$ is defined as: $$\operatorname{Li}_s(z)=\sum_{k=1}^\infty \frac{z^k}{k^s}$$ Plugging in $s=2$ yields $$\operatorname{Li}_2(z)=\sum_{k=1}^\infty \frac{z^k}{k^2}$$ This special case is called the Dilogarithm or Spence's function$$\\$$we can see that this is exactly the sum you were looking for,If we call it $S$, then $$S=\operatorname{Li}_2(x)$$ note that $S$ only converges for $\lvert x\rvert\le1$