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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. I asked for inductive proof so that I could use induction steps in the above inequality. But I did not get one.

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For the people who are interested, inspired from this answer, $$ \begin{align} s_m(n) &= \sin\left(\frac{t}{2}\right)\sum_{k=0}^nf_m(k)\sin(k+0.5)t\\ &=\frac{1}{2}\sum_{k=0}^nf_m(k)2\sin(k+0.5)t\sin\left(\frac{t}{2}\right)\\ &=\frac{1}{2}\sum_{k=0}^nf_m(k)[\cos kt -\cos(k+1)t]\\ &=\frac{1}{2}\left[f_m(0)[1 -\cos t]+...+f_m(n)[\cos nt -\cos(n+1)t]\right]\\ &=\frac{1}{2}\left[f_m(0) - f_m(n)\cos(n+1)t + [f_m(1)-f_m(0)]\cos t +...+[f_m(n)-f_m(n-1)]\cos nt\right] \end{align} $$

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.

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