The question is closely connected to another question of mine (see Show $\lim\limits_{x\to0}x\ln(x)=0$ by using Taylor series) but it emphasizes rather on the Taylor series itself.
Let's assume we already know that $\ln(x)=\sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{(x-1)^k}{k}$ for all $x\in(0,2]$. In other words $\ln(x)$ is the limit of its Taylor series expanded at point $a=1$.
How do I show that $x\ln(x)$ is the limit of its Taylor series expanded at $a=1$, $x\ln(x)\overset{??}{=}(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ for all $x\in[0,2]$?
My approach:
Let be $$ T_n(x):=(x-1)+\sum\limits_{k=2}^{n}(-1)^{k}\frac{(x-1)^k}{(k-1)k},\\ S_n(x):=1+\sum\limits_{k=1}^{n}(-1)^{k+1}\frac{(x-1)^k}{k}. $$ Then we see quickly that $T(x):=\lim\limits_{n\to\infty}T_n(x)$ exists for all $x\in[0,2]$ and that $(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ converges uniformly on $[0,2]$ (I won't go into further details because this is clear to me). Moreover we know that $1+\sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{(x-1)^k}{k}$ converges uniformly on $[\delta,2]$, where $0<\delta$.
If we take the derivative of $T_n(x)$ then we see that $T'_n(x)=S_{n-1}(x)$ and it follows that (see https://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability) $T'(x):=\lim\limits_{n\to\infty}T'_n(x)=\lim\limits_{n\to\infty}S_n(x)=1+\ln(x)$ for all $x\in(\delta,2]$. Additionally, it follows that $T_n(x)$ converges to the antiderivative of $\lim\limits_{n\to\infty}T'_n(x)=S_n(x)=1+\ln(x)$ which is equal to $x\ln(x)$. So $x\ln(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ on $[\delta,2]$. The case when $x=0$ needs to be handled separately: $(0-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(0-1)^k}{(k-1)k}=0=0\ln(0)$. Hence, $x\ln(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ on the whole intervall $[0,2]$.
Is this correct? Any comments or suggestions are welcome :)
Your question was answered, already (and you gave a link to the answer): for $|x|<1$, you can get the Taylor series of $x\,\ln x$ as the product of the Taylor series $x=1+(x-1)$ and $\ln(1+(x-1)))$, or by integration of $1+\ln(1+(x-1)))$. For $|x|=1$, that follows from Abel's theorem. If you have problems with those arguments, you should address them specifically, and not just restate the question, risking to receive equally unintelligible (for you) answers.