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.

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