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Is the function $$f(x)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\ldots\infty$$ continuous $\forall x\in\mathbb{R}$? I think no, probably the point of discontinuity is 0. Am I right in that? I think we should use the sum of a geometric series here. Also, is there any relationship to concepts of uniform convergence, like dini theorem etc.? Any hints? Thanks beforehand.

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Since $f(0)=0$ and since, if $x\neq 0$,$$x^2\left(1+\frac1{1+x^2}+\left(\frac1{1+x^2}\right)^2+\cdots\right)=x^2\frac1{1-\frac1{1+x^2}}=1+x^2,$$your function is discontinuous at $0$, since $\lim_{x\to0}f(x)=1\neq0=f(0)$.

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  • $\begingroup$ yes, I expected this, but is there any relation to uniform convergence like dini theorem etc.? $\endgroup$ – vidyarthi Jan 9 '18 at 12:58
  • $\begingroup$ @vidyarthi You can deduce from what I wrote that the convergence is not uniform, since you have a series of continuous functions which converge pointwise to a discontinuous one. $\endgroup$ – José Carlos Santos Jan 9 '18 at 12:59

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