Is this sequence defined correctly?

Question

I have some expression to develop, and this result reflects the brute force I used to prove that expression, however, I don't know whether that sequence is true or false.

$$\sum_{n=0}^{\infty} x^n \log (x^n) = \log\left(x^{x/(1-x)^2}\right) $$

Answer 1

HINT

At $n=0$ the sum term is 0, and $\ln(a^b) = b\ln a$ $$ \begin{split} \sum_{n=0}^\infty x^n \ln(x^n) &= \sum_{n=1}^\infty nx^n \ln x \\ &= x \ln x \sum_{n=1}^\infty nx^{n-1} \\ &= x \ln x \frac{d}{dx} \left[\sum_{n=0}^\infty x^n \right] \\ &= x \ln x \frac{d}{dx} \left[(1-x)^{-1}\right] \\ &= x (\ln x) (1-x)^{-2} \\ &= \ln \left( x ^{x/(1-x)^2}\right) \end{split} $$ but the geometric series expansion is only valid for $|x|<1$.

Answer 2

$$x^n\log x^n=nx^n\log x$$

and also $\;nx^n=x\left(x^n\right)'\;$ , so for $\;|x|<1\;$ , we get

$$\sum_{n=0}^\infty x^n=\frac1{1-x}\implies \left(\sum_{n=0}^\infty x^n\right)'=\sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}\implies$$

$$\sum_{n=0}^\infty x^n\log x^n=\log x\left(x\sum_{n=0}^\infty \left(x^n\right)'\right)=\log x\cdot \frac x{(1-x)^2}=\log x^{x/(1-x)^2}$$

so yes: your expression is correct... for $\;|x|<1\;$ . BTW, I'd rather leave the expression as $\;\cfrac{x\log x}{(1-x)^2}\;$ ...I think it is neater, not to mention that if $\;-1<x<0\;$ all the above is true...but then you mess up with complex logarithms