Taylor series for $f(x)=x\ln x$ at $x=1$

Question

I've been trying to figure out the Taylor series for the function $f(x)=x\ln(x)$ about $x=1$.

I know you can brute-force this and calculate some derivatives to get the final series, which is

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

However, I think you could just get the same result by multiplying the Taylor series for lnx about $x=1$ and x to get the series for xlnx. I know that the Taylor series for lnx is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}$, and I figured it would be best to use $x=(x-1)+1$.

With this, I get

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

which is nowhere near the series I want. What am I doing wrong?

Answer 1

With this, I get $x\ln(x)=((x-1)+1)\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n+1}}{n}+\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}$

Start from here,

$$x\ln(x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n+1}}{n}+\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}=A+B$$

Where

$$\begin{align}A&=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n+1}}{n},~~~\text{let}~~~m=n+1\\ \\ A&=\sum_{m=2}^{\infty}(-1)^{m}\frac{(x-1)^{m}}{m-1}\end{align}$$

For $B$, we write it as $$B=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}=(x-1)+\sum_{n=2}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}$$

Therefore,

$$\begin{align}A+B&=(x-1)+\sum_{n=2}^{\infty}\left(\frac{(-1)^{n}}{n-1}+\frac{(-1)^{n+1}}{n}\right)(x-1)^n\\ \\ &=(x-1)+\sum_{n=2}^{\infty}(-1)^{n}\left(\frac{1}{n-1}-\frac{1}{n}\right)(x-1)^n\\ \\ &=(x-1)+\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n(n-1)}\cdot (x-1)^n\end{align}$$

So we get:

$$\begin{align}x\ln x&=\sum_{n=1}^{\infty}a_n\cdot (x-1)^n\\ \\ a_1&=1,~~~a_n=\frac{(-1)^{n}}{n(n-1)},~~n=2, 3, 4...\end{align}$$

Answer 2

$$\begin{align}x\ln x&=\int_1^x(1+\ln t)dt\\&=x-1+\int_1^x\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(t-1)^n}ndt\\&=(x-1)+\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n+1}}{n(n+1)}\\&=(x-1)+\sum_{k=2}^{\infty}\frac{(-1)^k}{(k-1)k}(x-1)^k.\end{align}$$