Determine the Mclaurin series and the convergence radius for $(1-z)e^{-z}$.
I know the Taylor expansion of $e^t=\sum_\limits{n=0}^{\infty}\frac{t^n}{n!}$.Replacing $t=-z$ then I have $e^{-z}=\sum_\limits{n=0}^{\infty}\frac{(-1)^n z^n}{n!}$. Now the series would be: $(1-z)e^{-z}=(1-z)\sum_\limits{n=0}^{\infty}\frac{(-1)^n z^n}{n!}=...$
The books solution is $1+\sum_\limits{n=1}^{\infty}\frac{(-1)^{n+1}(n-1) z^n}{n!}$
I do not know how to manipulate the expression $(1-z)\sum_\limits{n=0}^{\infty}\frac{(-1)^n z^n}{n!}$ in order to get the solution.
Question:
Can someone help me solve the exercise?
Thanks in advance!
Your book's answer appears to be incorrect, or perhaps you've mistyped it.
It is perfectly accurate to say that $$(1-z)e^{-z}=(1-z)\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!},$$ but unfortunately, this isn't in the correct form, yet, since we need it to look like $$\sum_{n=0}^\infty a_nz^n$$ for some numbers $a_0,a_1,\dots.$ I'll walk you through how to get it there.
Note that for any $w\in\Bbb C,$ we have $(1-z)w=w-zw,$ so that $$(1-z)e^{-z}=\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!}-z\left(\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!}\right),$$ or equivalently, $$(1-z)e^{-z}=\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!}+(-1)z\left(\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!}\right).\tag{1}$$ Distributing the $(-1)z$ into the right-hand series in $(1),$ we obtain $$(1-z)e^{-z}=\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!}+\sum_{n=0}^\infty\frac{(-1)^{n+1}z^{n+1}}{n!},\tag{2}$$ but now our powers of $z$ don't match! The right-hand series in $(2)$ is adding up $$\frac{(-1)^{0+1}z^{0+1}}{0!}+\frac{(-1)^{1+1}z^{1+1}}{1!}+\frac{(-1)^{2+1}z^{2+1}}{2!}+\frac{(-1)^{3+1}z^{3+1}}{3!}+\cdots,$$ or equivalently, $$\frac{(-1)^1z^1}{(1-1)!}+\frac{(-1)^2z^2}{(2-1)!}+\frac{(-1)^3z^3}{(3-1)!}+\frac{(-1)^4z^4}{(4-1)!}+\cdots,$$ so we can make the substitution $n\mapsto n-1$ in the right-hand series in $(2)$ to get $$(1-z)e^{-z}=\sum_{n=0}^\infty\frac{(-1)^nz^n}{n!}+\sum_{n=1}^\infty\frac{(-1)^nz^n}{(n-1)!}.\tag{3}$$ Now the powers of $z$ in the two sums are matching, but the indices don't. However, we can simply take the $n=0$ term separately from the left-hand series in $(3),$ getting us $$(1-z)e^{-z}=1+\sum_{n=1}^\infty\frac{(-1)^nz^n}{n!}+\sum_{n=1}^\infty\frac{(-1)^nz^n}{(n-1)!},$$ at which point we can combine the sums to get $$\begin{eqnarray}(1-z)e^{-z} &=& 1+\sum_{n=1}^\infty\left(\frac{(-1)^nz^n}{n!}+\frac{(-1)^nz^n}{(n-1)!}\right)\\ &=& 1+\sum_{n=1}^\infty\left(\frac{(-1)^nz^n}{n!}+\frac{(-1)^nnz^n}{n!}\right)\\ &=& 1+\sum_{n=1}^\infty\frac{(-1)^n(n+1)z^n}{n!}.\end{eqnarray}$$ This is probably what your book's answer is (supposed to be). We can do even better, though! Noting that each of $(-1)^0,$ $0!,$ $(0+1)$ and $z^0$ are equal to $1,$ we then have $$1=\frac{(-1)^0(0+1)z^0}{0!},$$ and so we have $$(1-z)e^{-z}=\sum_{n=0}^\infty\frac{(-1)^n(n+1)}{n!}z^n.$$
Can you show that this series converges everywhere?