It doesn't look like anybody else is going to post a solution.
\( \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n \{ (2n+1)\sin^{2n+1}(k)+(-1)^n\cos^{2n+1}(k) \}}{\cos^{2n+1}(k)(2n+1)^2} \)
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Let \( \displaystyle S=\sum_{n=0}^{\infty}\frac{(-1)^n \{ (2n+1)\sin^{2n+1}(k)+(-1)^n\cos^{2n+1}(k) \}}{\cos^{2n+1}(k)(2n+1)^2} \)
\( \displaystyle \Rightarrow S=\sum_{n=0}^{\infty}\frac{(-1)^n (2n+1)\tan^{2n+1}(k)+1}{(2n+1)^2} \)
\( \displaystyle \Rightarrow S=\sum_{n=0}^{\infty}\left[\frac{(-1)^n \tan^{2n+1}(k)}{2n+1} +\frac{1}{(2n+1)^2} \right] \)
It is known that \( \displaystyle \arctan(z)=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{2n+1} \), so we get:
\( \displaystyle S= \arctan(\tan(k))+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} \)
\( \displaystyle \Rightarrow S= k+\left( \frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}\right) \)
\( \displaystyle \Rightarrow S= \boxed{\displaystyle k+\frac{\pi^2}{8}} \)
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\( \displaystyle \sum_{n=1}^{\infty} \frac{\sin^4(2^n)}{4^n} \)
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Note That: \(\displaystyle \frac{\sin^4(2^n)}{4^n}=\frac{\sin^2(2^n) \cdot \sin^2(2^n)}{4^n}= \frac{\sin^2(2^n) \cdot (1-\cos^2(2^n))}{4^n}= \frac{\sin^2(2^n)}{4^n}-\frac{\sin^2(2^{n+1})}{4^{n+1}}\)
\( \displaystyle \sum_{n=1}^{\infty} \frac{\sin^4(2^n)}{4^n} = \sum_{n=1}^{\infty} \frac{\sin^2(2^n)}{4^n}-\frac{\sin^2(2^{n+1})}{4^{n+1}}=\frac{\sin^2(2)}{4} +\sum_{n=2}^{\infty} \frac{\sin^2(2^n)}{4^n}-\sum_{n=1}^{\infty}\frac{\sin^2(2^{n+1})}{4^{n+1}} = \frac{\sin^2(2)}{4}+0=\boxed{\displaystyle \frac{\sin^2(2)}{4}}\)
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\( \displaystyle \sum_{n=0}^{\infty}\frac{\{ n+(n-1)x^2\}^2}{n!}\cdot \frac{x^{2n}}{(1+x^2)^{n+2}}\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)
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Let \( \displaystyle S=\sum_{n=0}^{\infty}\frac{\{ n+(n-1)x^2\}^2}{n!}\cdot \frac{x^{2n}}{(1+x^2)^{n+2}}\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)
\( \displaystyle \Rightarrow S=\sum_{n=0}^{\infty}\frac{\left\{ n - \frac{x^2}{1+x^2} \right\}^2}{n!}\cdot \left(\frac{x^{2}}{1+x^2} \right)^n\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)
Substitute \( \displaystyle y=\frac{x^2}{1+x^2} \).
\( \displaystyle \begin{align*} S &=e^{-y}\sum_{n=0}^{\infty}\frac{y^n(n-y)^2}{n!} \\ &=e^{-y}\sum_{n=0}^{\infty}\frac{n^2 y^n+y^{n+2}-2n y^{n+1}}{n!} \\ &= e^{-y} \left[\sum_{n=0}^{\infty}\frac{n^2 y^n}{n!}+ \sum_{n=0}^{\infty}\frac{ y^{n+2}}{n!}-\sum_{n=0}^{\infty}\frac{2n y^{n+1}}{n!}\right] \\ &= e^{-y} \left[y\sum_{n=0}^{\infty}\frac{n y^{n-1}}{(n-1)!}+ y^2\sum_{n=0}^{\infty}\frac{ y^{n}}{n!}-2y^2\sum_{n=0}^{\infty}\frac{ y^{n-1}}{(n-1)!}\right] \\ &= e^{-y} \left[y^2\sum_{n=0}^{\infty}\frac{ y^{n-2}}{(n-2)!}+y\sum_{n=0}^{\infty}\frac{ y^{n-1}}{(n-1)!}+ y^2e^y-2y^2 e^y\right]\\ &= e^{-y}(y e^y +2y^2 e^y-2y^2 e^y) \\&= e^{-y}(y e^y) \\&= y \end{align*}\)
\( \displaystyle S=y=\boxed{\displaystyle \frac{x^2}{1+x^2}} \).
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