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Question: Consider the series $g_n(x)=\sum_{k=0}^n\cfrac{x^2}{(1+x^2)^k}$ Prove that the series converges pointwise to the function $$g(x)=\begin{cases} 0 & \text{ if } x=0 \\ 1+x^2 & \text{ if } x \neq 0 \end{cases}$$ but the convergence is not uniform on any interval containing $0$ on its interior.

Help? Not even sure what the first step is on this one.

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You might first calculate the sum. Let $x\ne 0$. Then our series is a convergent geometric series. To check whether the convergence is uniform, look at the remainder if you stop summing at $m$. You will find a heavy dependence on $x$. Or else you can use properties of functions that are limits of uniformly convergent series. –  André Nicolas Dec 8 '12 at 3:39
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2 Answers

I'm assuming you're defining $$g_n(x)=\sum_{k=0}^n \frac{x^2}{(1+x^2)^k}$$

It is not hard to see that if $x=0$; $g_n(x)=0$ for each $n$ whence $g_n\to 0$. If $x\neq 0$, we then have that

$$ \lim \;g_n(x)=\sum_{k=0}^\infty \frac{x^2}{(1+x^2)^k}$$

$$ =x^2\sum_{k=0}^\infty \left(\frac{1}{1+x^2}\right)^k$$

$$ =x^2\frac{1}{1-\frac{1}{1+x^2}}$$

$$ =x^2\frac{1+x^2}{1+x^2-1}$$

$$ ={1+x^2}$$

and all the manipulations are justified for $$\frac 1 {1+x^2}<1$$ for any $x\neq 0$.

Can you see why the convergence is not uniform? If it were, your function would be continuous on an interval containing the origin.

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The property used by Peter Tamaroff is the best way to prove that we do not have uniform convergence. It can also be done directly from the definition.

Suppose that $x\gt 0$. Then the absolute value of the difference between the limit $\dfrac{1}{1+x^2}$ and the sum of the terms up to $\dfrac{x^2}{(1+x^2)^m}$ is $$\sum_{k=m+1}^\infty \frac{x^2}{(1+x^2)^k}.\tag{$1$}$$ This is an infinite geometric series. The usual formula gives that the sum $(1)$ is equal to $\dfrac{1}{(1+x^2)^m}$.

Let $\epsilon=1/2$. We show that there is no $N$ such that if $m\ge N$, then $(1)$ is $\lt \epsilon$ for every positive $x$. More informally, we show that there is no $N$ that "works" for every positive $x$.

Suppose to the contrary that there is such an $N$, and let $m=N$. Then $\dfrac{1}{(1+x^2)^N}\lt 1/2$. By algebraic manipulation, this is true precisely if $$|x|\gt \sqrt{2^{1/N}-1}.\tag{$2$}$$

From $(2)$, we see that there are non-zero $x$ for which the inequality does not hold, namely the positive $x$ that are $\le \sqrt{2^{1/N}-1}$. This contradicts the hypothesis of uniform convergence.

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