# Determining the uniform convergence

Show that the series ,whose partial sum of n terms is $$S_n=\frac{x}{(1+nx^2)}$$, converges uniformly for all real x.

I found that the series is pointwise convergent to 0 for all x. For showing uniform convergence, I found out that the function S attains maximum value at $$x=\frac{1}{\sqrt{n}}$$..i.e. $$M= \frac{1}{\sqrt{n}}$$.which tends to 0 as n tends to infinity.So it is proved that it is uniformly convergent.

However i am doubtful if this works for all real x or for particular closed interval.

Note the sum of the first $$n$$ terms as $$S_n(x)$$. We have
$$S_n(x) = \frac{x}{1+nx^2}$$ and $$\lim_{n \to \infty} S_n(x) = 0 \forall x \in \mathbb{R}.$$ Therefore the function $$S_n$$(x) converges point wise to $$0$$ on $$\mathbb{R}$$. To show uniform convergence to $$0$$ on $$D \subset \mathbb{R}$$ we must show that $$\forall \epsilon, \exists N \in \mathbb{N} : \sup_{x \in D}\vert S_n(x) - 0 \vert < \epsilon.$$
You have shown that $$\sup_{x \in \mathbb{R}} \vert S_n (x)\vert = S_n(\frac{1}{\sqrt{n}}) = \frac{1}{n + 1} \rightarrow 0 \text{ as } n \to \infty.$$
Remember by definition that $$\lim_{n \to \infty } a_n = L \in \mathbb{R} \iff \forall \epsilon , \exists N \in \mathbb{N},n > N : \vert a_n - L \vert < \epsilon$$. Therefore take $$a_n = \sup_{x \in \mathbb{R}} \vert S_n(x)\vert$$.
By definition $$S_n$$ converges uniformly to $$0$$ on $$\mathbb{R}$$.
Note that the function finally tends to $$0$$ any where. Therefore for uniform continuity we must show that$$\forall\epsilon>0\quad \exists N\quad\forall x\quad n>N\to|{x\over 1+nx^2}|<\epsilon$$where $$N=N(\epsilon)\ne N(\epsilon,x)$$. Also $$|{x\over 1+nx^2}|<\epsilon\iff 1+nx^2>{|x|\over \epsilon}\iff n>{1\over \epsilon |x|}-{1\over x^2}$$therefore by choosing $$N(\epsilon)=\max_{x\ne 0}{1\over \epsilon |x|}-{1\over x^2}={1\over 4\epsilon^2}$$ we obtain $$n>N(\epsilon)\to n>{1\over \epsilon |x|}-{1\over x^2}\to |S_n(x)|<\epsilon$$ and the proof is complete.