Laurent Series of $(z+1)^2\sin(z)$ centred at $0$ How am I to write the Laurent series of $(z+1)^2\sin(z)$ centred at $0$? I know
$$(z+1)^2 \sin(z) = (z^2+2z+1) \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!} \\  \Rightarrow (z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+3}}{(2k+1)!} + \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+2}}{(2k+1)!} + \sum_{k = 0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1)!},$$
but how am I to appropriately re-index is summation such that I might have a (closed-form) expression, containing only a single sum, as in, say,
$$(z+1)^2 \sin(z) = \sum_{k = 0}^{\infty} a_k z^{k}?$$
 A: We derive a series representation of $(z+1)^2\sin(z)$ evaluated at $z=0$ by calculating the coefficients $a_n$ in
\begin{align*}
\color{blue}{(z+1)^2\sin(z)}&=(z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\sum_{n=0}^{\infty}a_nz^n}\tag{1}\\
&=\sum_{n=0}^{\infty}a_{2n}z^{2n}+\sum_{n=0}^{\infty}a_{2n+1}z^{2n+1}\tag{2}\\
\end{align*}
At first we derive a representation (2) and then we merge the sums to obtain (1). We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. We obtain for $n\geq 1$:
\begin{align*}
\color{blue}{a_{2n}}&=[z^{2n}](z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&=\left([z^{2n-2}]+2[z^{2n-1}]+[z^{2n}]\right)\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\frac{2(-1)^{n-1}}{(2n-1)!}}\tag{3}\\
\color{blue}{a_{2n+1}}&=[z^{2n+1}](z+1)^2\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&=\left([z^{2n-1}]+2[z^{2n}]+[z^{2n+1}]\right)\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}\\
&\,\,\color{blue}{=\frac{(-1)^{n-1}}{(2n-1)!}+\frac{(-1)^n}{(2n+1)!}}\tag{4}
\end{align*}
We derive from (3) and (4) for $n\geq 1$:
\begin{align*}
a_n&=\frac{1+(-1)^n}{2}\,\color{blue}{\frac{2(-1)^{\frac{n}{2}-1}}{(n-1)!}}
+\frac{1-(-1)^n}{2}\left(\color{blue}{\frac{(-1)^{\frac{n-1}{2}-1}}{(n-2)!}+\frac{(-1)^{\frac{n-1}{2}}}{n!}}\right)\\
&=\frac{\left(1+(-1)^n\right)(-1)^{\frac{n}{2}-1}}{(n-1)!}
+\frac{\left(1-(-1)^n\right)(-1)^{\frac{n-1}{2}}}{2n!}\left(1+n-n^2\right)\tag{5}\\
\end{align*}
Comment:

*

*In (3) we select the coefficient of $z^{2n-1}$. The other coefficients with even power of $z$ are zero.


*In (4) we select the coefficients of $z^{2n-1}$ and $z^{2n+1}$. The coefficient with even power of $z$ is zero.


*In (5), we combine even and odd indexed coefficients $a_n$ by multiplying them by $\frac{1+(-1)^n}{2}$ and $\frac{1-(-1)^n}{2}$, respectively, and adding them.

We finally derive from (5)
\begin{align*}
&\color{blue}{(z+1)^2\sin(z)}\\
&\qquad=1+\sum_{n=1}^{\infty}\left(\left(1+(-1)^n\right)(-1)^{\frac{n}{2}-1}n\right.\\
&\qquad\qquad\qquad\qquad\left.+\left(1-(-1)^n\right)(-1)^{\frac{n-1}{2}}\frac{1+n-n^2}{2}\right)\frac{z^n}{n!}\\
&\qquad\color{blue}{=1+\sum_{n=1}^{\infty}\left(\left((-1)^{n-1}-1\right)n
-i\left(1-(-1)^n\right)\frac{1+n-n^2}{2}\right)\frac{\left(iz\right)^n}{n!}}\\
\end{align*}

