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:
$$s(n) = \sin\left(\frac{t}{2}\right)\sum\_{k=0}^n(k+1)\sin(k+0.5)t,\qquad t\in[0,\pi].$$



Inductive proof of $s(n)≤n+1$
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.


