<s>[INST] Taylor Series of $2xe^x$ I have to find the Taylor Series for $2xe^x$ centred at $x=1$. I came up with the following.
$$e^x = e^{x-1} \times e = e \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg)$$
Then consider $2xe^x$.
$$2xe^x = 2xe \bigg( \sum_{n=0}^\infty \frac{(x-1)^n}{n!}\bigg) = \bigg( \sum_{n=0}^\infty \frac{2xe(x-1)^n}{n!}\bigg) $$
I am wondering whether this is a right Taylor Series centred at $x=1$. I do understand that the $(x -1)^n$ implies that it is centred at $1$. But, I am very not sure with my answer since there are two terms of $x$, that is $x$ and $(x-1)^n$. 
Any clarification and explanation would be highly appreciated.
 [/INST] Write $2xe^x=2(x-1)e^x+2e^x$ and $e^x=e\cdot e^{x-1}$ to obtain $2xe^x=2e(x-1)e^{x-1}+2e\cdot e^{x-1}$. Since
$$
e^{x-1}=\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k,
$$
we have
\begin{align*}
2xe^x&=2e(x-1)e^{x-1}+2e\cdot e^{x-1} \\ &= 2e(x-1)\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k+2e
\sum_{k=0}^\infty\frac{1}{k!}(x-1)^k \\ &=
2e\left(\sum_{k=1}^\infty\frac{1}{(k-1)!}(x-1)^k+\sum_{k=1}^\infty\frac{1}{k!}(x-1)^k+1\right) \\ &=
2e\left(1+\sum_{k=1}^\infty\left(\frac{1}{(k-1)!}+\frac{1}{k!}\right)(x-1)^k\right) \\ &=
2e\sum_{k=0}^\infty\frac{k+1}{k!}(x-1)^k.
\end{align*}
</s>