deriving the sum of $x^n/(n+2)^2$

Question

I am writing a research paper and I have stumbled upon an issue.

I have to evaluate

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

Here is what I did:

$$ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$

$$\sum_{n=1}^{\infty} x^{n+1} = \frac{x^2}{1-x}$$

Integrate once with respect to $x$.

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

Divide by $x$

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

Integrate the expression again;

$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{(n+2)^2} = \frac{3\log(x)}{2} - x - \frac{x^2}{4} + Li_2(x)$$

All we have to do is divide by $x^2$

$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{3\log(x)}{2x^2} - \frac{1}{x} - \frac{1}{4} + \frac{Li_2(x)}{x^2} = \frac{4Li_2(x) - x^2 - 4x + 6\log(x)}{4x^2}$$

The issue is WolframAlpha returns it as:

$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{4Li_2(x) - x^2 - 4x}{4x^2}$$

Why is the way I did, wrong? I have an extraneous, extra log term.?

Help is appreciated.

Answer 1

Not sure what you did, but all that work seems kinda silly to arrive at your result.

$$\sum_{n=1}^\infty\frac{x^n}{(n+2)^2}=\frac{1}{x^2}\sum_{n=1}^\infty\frac{x^{n+2}}{(n+2)^2}=\frac{1}{x^2}(-x-\frac{x^2}{4}+\sum_{n=1}^\infty\frac{x^n}{n^2})=\frac{1}{x^2}(-x-\frac{x^2}{4}+\text{Li}_2(x))$$

Answer 2

Somewhere in your question, you write "Integrate once with respect to $x.$". In the equation following that phrase, you have a $\frac{3}{2}$ on the r.h.s. That's superfluous. The derivatives of the l.h.s. and the r.h.s are equal, and setting $x=0,$ we find that the constant of integration is $0.$