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Let $$F(x)=\sum_{n=1}^\infty \frac{(nx)}{n^2},$$ where $(x)=x$ for $x\in (-1/2,1/2]$ and $(x)$ is continued to $\mathbb{R}$ by periodicity, that is $(x+1)=(x)$. Prove that $F$ is discontinuous whenever $x=m/2n$, where $m,n\in \mathbb{Z}$ with $m$ odd and $n\neq 0$. This does not seem obvious to me.

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Isn't this just the difference between $\sum -n^{-2}$ vs. $\sum n^{-2}$? – user4143 Dec 3 '10 at 21:43
It suffices to find a sequence x_k such that x_k converges to m/2n but F(x_k) does not converge to F(m/2n). Think about how you would do this. – Qiaochu Yuan Dec 3 '10 at 21:45
up vote 1 down vote accepted

Note that $\lim_{x \rightarrow a}(x) = a$ except when $a$ is an odd multiple of ${1 \over 2}$, in which case $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 1$. Another way of saying this is that $\lim_{x \rightarrow a^-}(x) - \lim_{x \rightarrow a^+}(x) = 0$ unless $a$ is an odd multiple of ${1 \over 2}$ in which case it is 1.

So for the term $f_n(x) = {(nx) \over n^2}$, $\lim_{x \rightarrow a^-}f_n(x) - \lim_{x \rightarrow a^+}f_n(x) = 0$ unless $x = {m \over 2n}$ for some odd integer $m$, in which case it is ${1 \over n^2}$. So for your arbitrary ${m \over 2n}$ with $m$ odd, it will incur a jump of ${1 \over n^2}$ due to $f_n(x)$, and other terms can only increase or keep the same the size of the jump. For any $\epsilon > 0$, if enough terms have been added, the absolute value of the sum of the remaining terms will be at most $\epsilon$, so that no matter how these remaining terms add up there will still be a jump of at least ${1 \over n^2} - \epsilon$ at ${m \over 2n}$. Letting $\epsilon \rightarrow 0$ this means the infinite sum has a jump of at least ${1 \over n^2}$ at ${m \over 2n}$. (By jump I now mean the $\liminf$ from the left minus the $\limsup$ from the right.) In particular it's discontinuous there.

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You meant at $\frac{m}{2n}$ in the last two sentences? – TCL Dec 4 '10 at 1:17
yes, corrected that – Zarrax Dec 4 '10 at 1:44

A direct computation shows that $$F(x+)=F(x)-\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)-\frac{\pi^2}{16n^2},$$ $$F(x-)=F(x)+\frac{1}{2n^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=F(x)+\frac{\pi^2}{16n^2},$$ when $x=m/2n$, with $m$ and $2n$ being relatively prime. (The infinite sum is easily derived from the Euler result, of course.)

The function $F(x)$ was suggested by Riemann in his thesis as an example of a (Riemann)-integrable function which has a point of discontinuity in every interval (see "Gesammelte Mathematische Werke und Wissenschaftlicher Nachlass", Springer -Verlag 1990, pp. 259-296).

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For your version do you need $(x) = 0$ at ${1 \over 2}$ to have the symmetric limits? I'm thinking for the way he wrote it $F(x-) = F(x)$ and $F(x+) = F(x) - {\pi \over 16n^2}$. – Zarrax Dec 4 '10 at 16:05
Correction: $F(x-) = F(x)$ and $F(x+) = F(x) - {\pi \over 8n^2}$ – Zarrax Dec 4 '10 at 16:23

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