Commit 6f353c8e authored by Theo von Arx's avatar Theo von Arx

Add Sum k p^k

parent 1ab75605
......@@ -5,6 +5,7 @@
\item $\displaystyle \sum_{k=0}^\infty aq^k=a+aq+aq^2+\ldots=\frac{a}{1-q}$ für $|q|<1$
\item $\displaystyle \sum_{k=0}^\infty (k+1)q^k=1+2q+3q^2+\ldots=\frac{1}{(1-q)^2} $ für $|q|<1$
\item $\displaystyle \sum_{k=1}^\infty k q^k=\frac{q}{(1-q)^2} $ für $|q|<1$
\item $\displaystyle \sum_{k=1}^\infty \frac{1}{k^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots=\frac{\pi^2}{6}$
\item $\displaystyle \sum_{k=0}^\infty \frac{1}{k!}=1+1+\frac{1}{2}+\frac{1}{6}+\ldots=e$
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment