<s>[INST] Taylor series of $\sinh{(x)}$ at $\ln{(2)}$. Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
 Equating each derivative at $x = \ln{(2)}$ gives:
\begin{equation*}
\begin{split}
f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f''(\ln{(2)}) &=  \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(3)}(\ln{(2)}) &=  \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f^{(4)}(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(5)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4}
\end{split}
\end{equation*}
and so on. This means the Taylor series is
\begin{equation*}
\begin{split}
\sum_{n=0}^{\infty} \frac{f^{(n)}(\ln{(2)})}{n!}(x-\ln{(2)})^n &= f(\ln{(2)}) + \frac{f'(\ln{(2)})}{1!}(x-\ln{(2)}) + \frac{f''(\ln{(2)})}{2!}(x-\ln{(2)})^2 \\
&+ \frac{f^{(3)}(\ln{(2)})}{3!}(x-\ln{(2)})^3 + \frac{f^{(4)}(\ln{(2)})}{4!}(x-\ln{(2)})^4 + \frac{f^{(5)}(\ln{(2)})}{5!}(x-\ln{(2)})^5 + \ldots \\
&= \frac{3}{4}+\frac{5}{4\cdot1!}(x-\ln{(2)})^1+\frac{3}{4\cdot 2!}(x-\ln{(2)})^2 \\
&+\frac{5}{4\cdot 3!}(x-\ln{(2)})^3+\frac{3}{4\cdot 4!}(x-\ln{(2)})^4+\frac{5}{4\cdot 5!}(x-\ln{(2)})^5+\ldots
\end{split}
\end{equation*}
How do I find the general sum of this? Has it got something to do with the Maclaurin series which is $\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$.
 [/INST] $$\sinh(x)=\dfrac{e^x-e^{-x}}2=\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)!}$$
Set $x=\ln2$
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