Writing a sum in terms of an appropriate function I have a solution that is expressed as a series:
$$
\sum_{k=0}^{\infty}\left[\frac{(-1)^k t^{2k+1}}{(2k+1)!}\right]\left[4^k\right]
$$
.. and would like to show it in terms of an appropriate function, for instance either in terms of $\sin(t), \cos(t), e^t$, etc.
What is frustrating me is that if only the multiplier (first bracket) were in the sum, as opposed to both
the multiplier and the multiplicand (second bracket), I could express the sum as $\sin(t)$. However, since they are both in the sum, I cannot find an appropriate function. Does an appropriate function exist, and if so, what is it, and how did you find it?
EDIT: It seems $\tanh(x)$ is pretty close match but the coefficients of the even terms are slightly off
 A: You can write
$$ \sum_{k=0}^{\infty}\left[\frac{(-1)^k t^{2k+1}}{(2k+1)!}\right]\left[4^k\right]= 
\frac 12\sum_{k=0}^{\infty}\left[\frac{(-1)^k (2t)^{2k+1}}{(2k+1)!}\right]=\frac 12\sin(2t)$$
A: HINT
$$
4^k = 2^{2k} = \frac12 2^{2k+1}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}
\bracks{{\pars{-1}^{k}t^{2k + 1} \over
\pars{2k + 1}!}}\bracks{4^{k}}} =
-{1 \over 2}\,\ic\sum_{k = 0}^{\infty}{\ic^{2k + 1}\, t^{2k + 1} \over
\pars{2k + 1}!}\,2^{2k + 1}
\\[5mm] = &\
-{1 \over 2}\,\ic\sum_{k = 0}^{\infty}{\pars{2\ic t}^{k} \over
k!}\,{1 - \pars{-1}^{k} \over 2} =
-{1 \over 2}\,\ic\bracks{\sum_{k = 0}^{\infty}
{\pars{2\ic t}^{k} \over k!} -
\sum_{k = 0}^{\infty}{\pars{-2\ic t}^{k} \over k!}}
\\[5mm] = &\
-{1 \over 2}\,\ic\pars{\expo{2\ic t} - \expo{-2\ic t}} =
\bbx{\sin\pars{2t}} \\ &
\end{align}
