Check the uniform convergence of $\displaystyle \sum_{k=1}^{\infty} \left ( \frac{x \sin x}{1+x} \right )^k$ where $x \in [0,1]$.
Solution i tried - The sequence of function has a contiunous limit over given interval because $$\left ( \frac{x \sin x}{1+x} \right )^k \to 0\;\; \forall \;x \in [0,1]$$ after that i tried to take it as $$\left | \left ( \frac{x \sin x}{1+x} \right )^k \right | <\left ( \frac{1}{2} \right )^k$$ thus here given $\displaystyle \sum \left ( \frac{1}{2} \right )^k$ is a sum of G.P with common ration $\frac{1}{2}$ then it will be convergent so the given series is uniformaly convergent by $M_n$ test ,is every thing ok in my solution
Please help
Thank you