From 45d95c45013bf4ff73570c94c58b7f0212ccdf26 Mon Sep 17 00:00:00 2001
From: mzuenni <michi.zuendorf@gmail.com>
Date: Sun, 8 Sep 2024 22:16:36 +0200
Subject: moved stuff

---
 content/math/tables/series.tex | 32 ++++++++++++++++----------------
 1 file changed, 16 insertions(+), 16 deletions(-)

(limited to 'content/math/tables/series.tex')

diff --git a/content/math/tables/series.tex b/content/math/tables/series.tex
index 3042781..9618c2b 100644
--- a/content/math/tables/series.tex
+++ b/content/math/tables/series.tex
@@ -1,33 +1,33 @@
-\begin{tabularx}{\linewidth}{|XIXIXIX|}
-	\hline
-	\multicolumn{4}{|c|}{Reihen} \\
+\begin{expandtable}
+\begin{tabularx}{\linewidth}{|XIXIX|}
 	\hline
 	$\sum\limits_{i = 1}^n i = \frac{n(n+1)}{2}$ &
 	$\sum\limits_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$ &
-	$\sum\limits_{i = 1}^n i^3 = \frac{n^2 (n + 1)^2}{4}$ &
-	$H_n = \sum\limits_{i = 1}^n \frac{1}{i}$ \\
+	$\sum\limits_{i = 1}^n i^3 = \frac{n^2 (n + 1)^2}{4}$ \\
 	\grayhline
 	
-	$\sum\limits_{i = 0}^n c^i = \frac{c^{n + 1} - 1}{c - 1} \quad c \neq 1$ &	
-	$\sum\limits_{i = 0}^\infty c^i = \frac{1}{1 - c} \quad \vert c \vert < 1$ &
-	$\sum\limits_{i = 1}^\infty c^i = \frac{c}{1 - c} \quad \vert c \vert < 1$ &	
-	$\sum\limits_{i = 0}^\infty ic^i = \frac{c}{(1 - c)^2} \quad \vert c \vert < 1$ \\
+	$\sum\limits_{i = 0}^n c^i = \frac{c^{n + 1} - 1}{c - 1} \hfill c \neq 1$ &	
+	$\sum\limits_{i = 0}^\infty c^i = \frac{1}{1 - c} \hfill \vert c \vert < 1$ &
+	$\sum\limits_{i = 1}^\infty c^i = \frac{c}{1 - c} \hfill \vert c \vert < 1$ \\
 	\grayhline
-	
+		
 	\multicolumn{2}{|lI}{
 		$\sum\limits_{i = 0}^n ic^i = \frac{nc^{n + 2} - (n + 1)c^{n + 1} + c}{(c - 1)^2} \quad c \neq 1$
 	} &	
-	\multicolumn{2}{l|}{
+	$\sum\limits_{i = 0}^\infty ic^i = \frac{c}{(1 - c)^2} \hfill \vert c \vert < 1$ \\
+	\grayhline	
+	
+	\multicolumn{2}{|lI}{
 		$\sum\limits_{i = 1}^n iH_i = \frac{n(n + 1)}{2}H_n - \frac{n(n - 1)}{4}$
-	} \\
+	} &
+	$H_n = \sum\limits_{i = 1}^n \frac{1}{i}$ \\
 	\grayhline
 	
 	\multicolumn{2}{|lI}{
-		$\sum\limits_{i = 1}^n H_i = (n + 1)H_n - n$
-	} &
-	\multicolumn{2}{l|}{
 		$\sum\limits_{i = 1}^n \binom{i}{m}H_i =
 		\binom{n + 1}{m + 1} \left(H_{n + 1} - \frac{1}{m  + 1}\right)$
-	} \\
+	} &
+	$\sum\limits_{i = 1}^n H_i = (n + 1)H_n - n$ \\
 	\hline
 \end{tabularx}
+\end{expandtable}
-- 
cgit v1.2.3

