# $\sum_{n=-\infty}^{+\infty}\frac{1}{z-n}$ converges uniformly to a $1-$ periodic meromorphic function.

Given a sequence of meromorphic function $(f_n)_n$ on a open set $\Omega\subset \Bbb{C}.$

How does it mean that $\sum_{n=-\infty}^{+\infty}f_n$ converges uniformly on every compact of $D?$

Does it mean that $\sum_{n=1}^\infty (f_n+f_{-n})$ converges uniformly on every compact?

I ask because I find an exercise when it's asking to prove the convergence (on every compact) of $\sum_{n=-\infty}^{+\infty}\frac{1}{z-n}$ to a $1-$periodic meromorphic function. If I guess correctly the definition we have $$\frac1{z-n}+\frac1{z+n}=\frac{2z}{z^2-n^2}.$$

If $A=\sup\{\vert z\vert:z\in K\}$ then $\vert\frac{2z}{z^2-n^2}\vert\le\frac{2A}{n^2-A^2}$ for $n>A.$ So that the serie converges uniformly and it's meromorphic by a theorem.

But why it's $1-periodic?$

• In principle, it means that $\sum_{n = 0}^{\infty} f_n$ and $\sum_{m = 1}^{\infty} f_{-m}$ both converge uniformly. But the given series does not converge in the default sense. It does converge (locally uniformly) in a weaker sense that considers only the symmetric partial sums $\sum_{n = -k}^k f_n$. That weaker sense is sufficient for most things you want to do, but you need to be more careful that your manipulations are legitimate than with the usual sense of convergence of a $\mathbb{Z}$-indexed series. – Daniel Fischer May 13 '17 at 13:07
• Ok I see. So $\sum_{n=0}^\infty f_n$ does not converge uniformly because it does not converge "pointwise"? And the definition given it's called locally uniformly ? Thanks. – Alex May 13 '17 at 13:17
• "Locally uniformly" is (since $\mathbb{C}$ is locally compact) the same as "uniformly on every compact set". [On spaces that aren't locally compact, locally uniform convergence implies uniform convergence on every compact subset, but not vice versa.] Indeed, since $\sum_{n = 0}^{\infty} \frac{1}{z-n}$ doesn't converge at any $z\in \mathbb{C}$, it a fortiori doesn't converge uniformly on any $S\subset \mathbb{C}$. – Daniel Fischer May 13 '17 at 13:27
• Ok I see, I didn't think about $\Bbb{C}$ being locally compact. Thank you. @DanielFischer – Alex May 13 '17 at 13:32

$\displaystyle\sum_{n=-\infty}^\infty \frac{1}{z-n}$ diverges but $\displaystyle\frac{1}{2}\sum_{n=-\infty}^\infty (\frac{1}{z-n}+\frac{1}{z+n})$ converges uniformly away from its poles. It is obviously $1$-periodic.

• Could you give a proof for uniform convergence of $\displaystyle\frac{1}{2}\sum_{n=-\infty}^\infty (\frac{1}{z-n}+\frac{1}{z+n})$ ? – Isa Aug 28 '18 at 19:57

is the sine function remember that $\frac{1}{z-n}+\frac{1}{z+n}$ is just $\frac{2z}{z^{2}-n^{2}}$ for every n , now integrate over z and you get

$$\sum_{n=0}^{\infty}ln(z^{2}-n^{2}$$

which is the logarithm of the sine functio divided by x $ln(sinx) -ln(x)$