Question

Prove the series sinx/1-sin2x/2^3+ sin3x/3^3.. Converges absolutely

Answer 1

Correct answer is $\left | S_{n} \right |$ $\leq \sum_{1}^{\infty}\frac{1}{n^{^{3}}}$

Explanation:

$\sum S_{n} = \frac{\sin x}{1^{3}} - \frac{\sin 2x}{2^{3}} + \frac{\sin 3x}{3^{3}} -...$

$\textrm{Absolute term of this series is}$

$\sum S_{n} = \frac{\begin{vmatrix}\sin x\end{vmatrix}}{1^{3}} + \frac{\begin{vmatrix}\sin 2x\end{vmatrix} }{2^{3}} + \frac{\begin{vmatrix}\sin 3x\end{vmatrix}}{3^{3}} +...$

$\textrm{As we know that}$  $1\leq \sin x\leq 1 \Rightarrow \begin{vmatrix}\sin x \end{vmatrix} \leq 1 \forall n \in N$

$\rightarrow \frac{\left | \sin x \right |}{n^{3}} \leq \frac{1}{n^{3}}$

$\therefore \sum \left | S_{n} \right |$$\leq \sum_{1}^{\infty}\frac{1}{n^{^{3}}}$