Inductive proof of $s(n)≤n+1$ I was able to conclude, numerically, the following:
$$s(n) = \sum_{j=0}^n\frac{(-4)^j}{(2j+1)!}\left(\sum_{k=j}^n\frac{(k+1)(2k+1)(k+j)!}{(k-j)!}\right)x^{2j+2}\le n+1$$
for $x\in[0,1]$. For example
\begin{array}{ccc}
   n& max(s(n)) & n+1 \\
   0&  1.00&  1\\
   1&  1.53&  2\\
   2&  2.07&  3\\
   3&  2.60&  4\\
   4&  3.13&  5\\
\end{array}
But I was unable to prove it analytically. My need is an analytical proof. An inductive proof is preferable yet other proofs are also welcome.
P. S.: The $s(n)$ has an alternate form as following:
$$s(n) = \sin\left(\frac{t}{2}\right)\sum_{k=0}^n(k+1)\sin(k+0.5)t,\qquad t\in[0,\pi].$$
 A: The alternative form can be expressed as
$$ \frac{1}{2} - \frac{n+1}{2} \cos (n+1)t 
+ \frac{1}{2}\bigl( \cos nt + \cdots + \cos t \bigr).$$
The third summand has period $2\pi$ and takes its maximum value of $n/2$ when $t=0$. So just from the triangle inequality we get
$$s(n) \le \frac{1}{2} + \frac{n+1}{2} + \frac{n}{2} = n+1.$$ 
This answers your question. I expect that the Fourier series above can be used to get an even stronger inequality.
A: For $t\in[0,\pi]$,
\begin{align}
s(n)&=\sin(t/2)\sum_{k=0}^n(k+1)\Im(e^{\mathrm{i}(k+1/2)t})\\
&=\Im\left(e^{-\mathrm{i}t/2}\sin(t/2)\sum_{k=0}^n(k+1)e^{\mathrm{i}(k+1)t}\right)\\
&=\Im\left(e^{-\mathrm{i}t/2}\frac{e^{\mathrm{i}t/2}-e^{-\mathrm{i}t/2}}{2\mathrm{i}}\frac{1}{\mathrm{i}}\frac{\,d}{\,dt}\frac{e^{\mathrm{i}t}-e^{\mathrm{i}(n+2)t}}{1-e^{\mathrm{i}t}}\right)\\
&=\frac{1}{2}\Im\left((1-e^{-\mathrm{i}t})\frac{\,d}{\,dt}\frac{1-e^{\mathrm{i}(n+1)t}}{1-e^{-\mathrm{i}t}}\right)\\
&=\frac{1}{2}\Im\left(\mathrm{i}\frac{1-e^{\mathrm{i}(n+1)t}}{1-e^{\mathrm{i}t}}-\mathrm{i}(n+1)e^{\mathrm{i}(n+1)t}\right)\\
&=\frac{1}{4}\frac{\sin((n+1/2)t)+\sin(t/2)}{\sin(t/2)}-\frac{n+1}{2}(\cos(n+1)t).
\end{align}
Thus,
\begin{align}
s(n)&\le \frac{1}{4}\frac{\sin((2n+1)t/2)}{\sin(t/2)}+\frac{1}{4}+\frac{n+1}{2}\\
&=\frac{1}{4}\sum_{k=-n}^{n}e^{\mathrm{i}kt}+\frac{1}{4}+\frac{n+1}{2}\le \frac{2n+1}{4}+\frac{1}{4}+\frac{n+1}{2}= n+1.
\end{align}
