Uniform convergence: $\sum \frac{x^2}{(1 + x^2)^n}$

Question

Consider the series of functions:

$$\sum_{n \ge 1} \frac{x^2}{(1 + x^2)^n}$$

Q: Where does this series converge uniformly?

We have if $x \neq 0$:

$$\lim_{n \to \infty} \left| \frac{x^2}{(1 + x^2)^{n+1}} / \frac{x^2}{(1 + x^2)^n} \right|= \frac{1}{1 + x^2} < 1$$

And if $x = 0$, then the series converges. Therefore, we have pointwise convergence everywhere.

For uniform convergence:

$$R_n(x) = \sum_{k = n+1}^{\infty} \frac{x^2}{(1 + x^2)^k} \ge \sum_{k = n+1}^{2n} \frac{x^2}{(1 + x^2)^k} \ge \sum_{k = n+1}^{2n} \frac{x^2}{(1 + x^2)^{2n}} = \frac{nx^2}{(1 + x^2)^{2n}}, \ \forall x \in \mathbb {R}$$

Then:

$$R_n(\frac{1}{\sqrt{n}}) \ge \frac{1}{(1 + 1/n)^n)^2} \longrightarrow \frac{1}{e^2} > 0$$

So, we can't have uniform convergence in a neighbourhood of $0$.

It seems that for $a > 0$, we don't have uniform convergence on $[a, \infty)$ or $(-\infty, -a]$ because we can make $x$ as large as we desire. But, I'm not sure about this.

However, on any compact interval $[a,b]$ with $b < -1$ or $a > 1$, we do have uniform convergence because of normal convergence:

$$\sup_{x \in [a,b]} \left( \frac{x^2}{(1 + x^2)^n} \right) \le \frac{b^2}{a^{2n}}$$

Can someone help me find all the sets on which we have uniform convergence?

Thank you.

Answer 1

You can actually compute the partial sums of this series, which I expect will help: it's geometric with common ratio $1/(1+x^2)$ and first term $x^2$, so we have $$ \sum_{n=1}^m \frac{x^2}{(1+x^2)^n} = 1-\frac{1}{(1+x^2)^m}. $$ Hence it converges to $1$ when $x^2>0$ and $0$ when $x=0$. Therefore, since a uniform limit of continuous functions is continuous, the convergence cannot be uniform on any set containing $x=0$. It is easy to see, however, that the error after summing the first $m$ terms is $$ \frac{1}{(1+x^2)^m} < \frac{1}{(1+r^2)^m} \to 0 \quad \text{as } \, m \to \infty $$ for any $x$ with $\lvert x \rvert \geqslant r$. Since this works for any $r>0$, the convergence is uniform on sets of the form $ (-\infty,-r] \cap [r,\infty) $, $r>0$ (and hence on finite unions and intersections of these as well, in the obvious way).

Answer 2

Perhaps I am missing something, but I believe we can write your sequence as:

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

This gives two infinite sums:

\begin{align} \sum_1^\infty \frac{x^2}{(1 + x^2)^n} =& \sum_1^\infty \left(\frac{1}{1 + x^2}\right)^{n - 1} - \sum_1^\infty \left(\frac{1}{1 + x^2}\right)^n \\ =&\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n - \left(\frac{1}{1 + x^2}\right)\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \\ =& \left(1 - \left(\frac{1}{1 + x^2}\right)\right)\sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \\ =& \frac{x^2}{1 + x^2} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n \end{align}

Since the only sum is now a geometric series with $r = \frac{1}{1 + x^2}$ giving the sum (when $|x| > 0$):

\begin{align} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n = \frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1 + x^2}{x^2} \\ \sum_1^\infty \frac{x^2}{(1 + x^2)^n} = \frac{x^2}{1 + x^2} \sum_0^\infty \left(\frac{1}{1 + x^2}\right)^n = \frac{x^2}{1 + x^2} \frac{1 + x^2}{x^2} = 1 \end{align}

This series is identically $1$ everywhere except at $x = 0$ when the sum is clearly $0$. Therefore this function $f(x) = \sum_1^\infty \frac{x^2}{(1 + x^2)^n}$ is discontinuous at the value $x = 0$ but continuous and differentiable everywhere else.

Another way to see the above is that the original rearrangement gave a telescopic series:

$$ 1 - \frac{1}{1 + x^2} + \frac{1}{1 + x^2} - \frac{1}{(1 + x^2)^2} + \frac{1}{(1 + x^2)^2} ... - \frac{1}{(1 + x^2)^n} $$

If $|x| > 0$ then $1 + x^2 > 1$ and thus $\lim_{n \rightarrow \infty}\left(1 - \frac{1}{(1 + x^2)^n}\right) = 1$. Even when $x = 0$, for all $n$ we have: $1 - \frac{1}{1^n} = 0$ (which agrees with the observation that the sum is $0$ when $x = 0$).

Answer 3

On $[2,\infty)$ the $n$th summand is $< x^2/(x^2)^{n}\le 1/4^{n-1}.$ So the series converges uniformly on $[2,\infty).$ That suggests the series converges uniformly on every $\mathbb {R}\setminus U,$ where $U$ is a neighborhood of $0.$