For fixed $m = 0, 1, 2, ...$
$$f\_m(k) = \prod\_{j=1}^{m}(k+j).$$
Some examples of $f\_m(k)$ are as following:
$$f\_0(k) = 1, \quad f\_1(k) = (k+1), \quad f\_2(k) = (k+1)(k+2).$$


The $s\_m(n)$ is defined as following:
$$s\_m(n) = \sin\left(\frac{t}{2}\right)\sum\_{k=0}^nf\_m(k)\sin(k+0.5)t,\qquad t\in[0,\pi].$$


The $s\_m(n)$ can also be defined as following:
$$s\_m(n) = \sum\_{j=0}^n\frac{(-4)^j}{(2j+1)!}\left(\sum\_{k=j}^n\frac{f\_m(k)(2k+1)(k+j)!}{(k-j)!}\right)x^{2j+2},\qquad x\in[0,1].$$


I want to prove
$$|s\_m(n)| \le f\_m(n), \forall x ~or ~t$$


I am sure the inequality holds but I am unable to prove it. I used MATLAB and verified the inequality for some values of $m$ and $n$ as presented below:


\begin{array}{ccccccccc}
 n & \max(s\_0(n)) & f\_0(n) & \max(s\_1(n)) & f\_1(n) & \max(s\_2(n))& f\_2(n) & \max(s\_3(n)) & f\_3(n)\\
 0 & 1.00 & 1 & 1.00 & 1 & 2.00 & 2 & 6.00 & 6 \\
 1 & 1.00 & 1 & 1.53 & 2 & 4.17 & 6 & 18.00 & 24 \\
 2 & 1.00 & 1 & 2.07 & 3 & 8.00 & 12 & 42.00 & 60 \\
 3 & 1.00 & 1 & 2.60 & 4 & 12.46 & 20 & 78.30 & 120 \\
 4 & 1.00 & 1 & 3.13 & 5 & 18.03 & 30 & 132.00 & 210 
\end{array}


Any help will be greatly appreciated.


**PS**: Please refer to [this question](https://mathoverflow.net/questions/306118/inductive-proof-of-sn%E2%89%A4n1). I asked for inductive proof so that I could use induction steps in the above inequality. But I did not get one.



Proof of an inequality $s_m(n) \le f_m(n)$
For the people who are interested, inspired from [this answer](https://mathoverflow.net/a/306131/126661),
$$
\begin{align}
s\_m(n) 
&= \sin\left(\frac{t}{2}\right)\sum\_{k=0}^nf\_m(k)\sin(k+0.5)t\\
&=\frac{1}{2}\sum\_{k=0}^nf\_m(k)2\sin(k+0.5)t\sin\left(\frac{t}{2}\right)\\
&=\frac{1}{2}\sum\_{k=0}^nf\_m(k)[\cos kt -\cos(k+1)t]\\
&=\frac{1}{2}\left[f\_m(0)[1 -\cos t]+...+f\_m(n)[\cos nt -\cos(n+1)t]\right]\\
&=\frac{1}{2}\left[f\_m(0) - f\_m(n)\cos(n+1)t + [f\_m(1)-f\_m(0)]\cos t +...+[f\_m(n)-f\_m(n-1)]\cos nt\right]
\end{align}
$$


So
$$
\begin{align}
|s\_m(n)| 
&=\left|\frac{1}{2}\left[f\_m(0) - f\_m(n)\cos(n+1)t + [f\_m(1)-f\_m(0)]\cos t +...+[f\_m(n)-f\_m(n-1)]\cos nt\right]\right|\\
&\le \frac{1}{2}\left[f\_m(0) + f\_m(n)|\cos(n+1)t|+[f\_m(1)-f\_m(0)]|\cos t|+...+[f\_m(n)-f\_m(n-1)] |\cos nt|\right]\\
&\le \frac{1}{2}\left[f\_m(0) + f\_m(n)+[f\_m(1)-f\_m(0)]+...+[f\_m(n)-f\_m(n-1)] \right]\\
&=f\_m(n).
\end{align}
$$


I have used $f\_m(k)$ is positive and increasing.


