Hi folks,
Let $p>1$. I've to show that $\sum_{n=1}^\infty \frac{\pi}{2} - \arctan(n^p(1+x^2))$ converges uniformly in $\mathbb{R}$.
Is my approach ok?
1.Approach:
Let $\epsilon>0$ be given. Since $\arctan(x)$ converges to $\pi/2$ for $x \rightarrow \infty$ we can find some positive integer $N$ such that $|\frac{\pi}{2} -\arctan(n^p(1+x^2))|<\frac{\epsilon}{m-n+1} for all $m>n>N$. Thus $|\sum_{n=1}^\infty \frac{\pi}{2} - \arctan(n^p(1+x^2))|\leq \sum_{n=1}^\infty |\frac{\pi}{2} - \arctan(n^p(1+x^2))|<\epsilon$ for all $ m>n>N$.
Let $p>1$. I've to show that $\sum_{n=1}^\infty \frac{\pi}{2} - \arctan(n^p(1+x^2))$ converges uniformly in $\mathbb{R}$.
Is my approach ok?
1.Approach:
Let $\epsilon>0$ be given. Since $\arctan(x)$ converges to $\pi/2$ for $x \rightarrow \infty$ we can find some positive integer $N$ such that $|\frac{\pi}{2} -\arctan(n^p(1+x^2))|<\frac{\epsilon}{m-n+1} for all $m>n>N$. Thus $|\sum_{n=1}^\infty \frac{\pi}{2} - \arctan(n^p(1+x^2))|\leq \sum_{n=1}^\infty |\frac{\pi}{2} - \arctan(n^p(1+x^2))|<\epsilon$ for all $ m>n>N$.