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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 $$s_m(n) \le f_m(n).$$$$|s_m(n)| \le f_m(n), \forall x ~or ~t$$

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.

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 $$s_m(n) \le f_m(n).$$

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.

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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Proof of an inequaliyinequality $s_m(n) \le f_m(n)$

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Proof of an inequaliy $s_m(n) \le f_m(n)$

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 $$s_m(n) \le f_m(n).$$

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.