Unendliche Summen

\[ \begin{align} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2}}} &= \frac{\pi^2}{6} \\ \notag \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{4}}} &= \frac{\pi^4}{90} \\ \notag \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{6}}} &= \frac{\pi^6}{945} \\ \notag \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{8}}} &= \frac{\pi^8}{9450} \\ \notag \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{10}}} &= \frac{\pi^{10}}{93555} \\ \notag \\
\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{12}}} &= \frac{691\pi^{12}}{638512875} \\ \notag \\ \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2p}}} &= { \frac{(-1)^{p+1}p\,\pi^{2p}}{(2p+1)!} – \sum \limits_{q = 1}^{p-1}\frac{(-1)^{p+q}\pi^{2p-2q}}{(2p-2q+1)!} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2q}}}},\quad p \in \mathbb{N} \\ \notag \\  \frac{p\,\pi^{2p}}{(2p+1)!} &=  \sum \limits_{q = 1}^{p}\frac{(-1)^{p+q}\pi^{2p-2q}}{(2p-2q+1)!} \sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2q}}},\quad p \in \mathbb{N} \\ \notag \\ \sum \limits_{n = 1}^{\infty}{\frac{1}{1+r^2n^{2}}} &= \frac{1}{2} {\left[{\frac{\pi}{r}\coth{\left({\frac{\pi}{r}}\right)}  – 1}\right]},\quad r\in  \mathbb{R},\, r\ne{0} \\ \notag \\ \sum \limits_{n = 1}^{\infty}{\frac{(-1)^{n+1}}{(2n-1)^{3}}} &= \frac{\pi^{3}}{32} \\ \notag \\ \sum \limits_{n = 1}^{\infty}{\frac{(-1)^{n+1}}{(2n-1)^{5}}} &= \frac{5\pi^{5}}{1536} \\ \notag \\  \sum \limits_{n = 1}^{\infty}{\frac{1}{(2n-1)^{4}}} &= \frac{\pi^{4}}{96} \\ \notag \\  \sum \limits_{n = 1}^{\infty}{\frac{1}{(2n-1)^{6}}} &= \frac{\pi^{6}}{960} \\ \notag \\ \sum \limits_{n = 1}^{\infty}{\frac{1}{(2n-1)^{2p}}} &= \frac{2^{2p-1}-1}{2^{2p}}\sum \limits_{n = 1}^{\infty}{\frac{1}{n^{2p}}},\quad p \in \mathbb{N} \\ \notag \\  \sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)}}&= 1 \\ \notag \\  \sum\limits_{n=1}^{\infty}\frac{\left\lfloor{\sqrt{n}}\right\rfloor}{n(n+1)} &= \frac{{\pi}^2}{6} \\ \notag \\  \sum\limits_{n=1}^{\infty}\frac{\left\lfloor{n^{\frac{1}{4}}}\right\rfloor}{n(n+1)} &= \frac{{\pi}^4}{90} \\ \notag \\  \sum\limits_{n=1}^{\infty}\frac{\left \lfloor {n^\frac{1}{p}} \right\rfloor}{n(n+1)} &= \sum\limits_{n=1}^{\infty} \frac{1}{{n}^{p}}, \quad {p}\in \mathbb{R}, \,p\gt{1}  \\ \notag \\  \sum\limits_{n=1}^{\infty}\frac{\left\lfloor{n^{\frac{1}{2}}}\right\rfloor}{n(n+1)(n+2)} &= \frac{3 + {\pi}^2\, -\, {3\,\pi} \coth \pi}{12} \\ \notag \\  \sum\limits_{n=1}^{\infty}\frac{\left \lfloor {n^\frac{1}{p}} \right\rfloor}{n(n+1)(n+2)} &= \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{1}{{n}^{p}{\left({{n}^{p} + 1}\right)}}, \quad {p}\in \mathbb{R}, \,p\gt{1}  \\ \notag \\  \sum\limits_{n=1}^{\infty}{\frac{p!}{n(n+1)(n+2)\cdots (n+p)}} &= \sum\limits_{l=1}^{p}{\frac{1}{l}\sum\limits_{j=0}^{l-1}{\binom{p}{j}(-1)^j}}, \quad p\in \mathbb{N}  \\ \notag \\ \sum\limits_{n=p}^{\infty}{\frac{1}{\binom{n+1}{p+1}}} &= (p+1)\sum\limits_{l=1}^{p}{\sum\limits_{j=0}^{l-1}{\binom{p}{j}\frac{(-1)^j}{l}}}, \quad p\in \mathbb{N}  \\ \notag \\ \sum\limits_{n=0}^{\infty} {{\left \lfloor {\frac{n}{p}} \right\rfloor}x^n} &=  \frac{x^p}{(1-x)\left(1-x^p\right)}, \quad p\in \mathbb{R}, p\ge{1},\mid{x}\mid\lt{1} \\ \notag \\ \sum\limits_{n=p}^{\infty} {\frac{x^n}{\left \lfloor {\frac{n}{p}} \right\rfloor}} &=  -\frac{1-x^p}{1-x}{\log\left(1-x^p\right)}, \quad p\in \mathbb{R}, p\ge{1},\mid{x}\mid\lt{1} \\ \notag \\  \sum\limits_{n=1}^{\infty}\frac{\left \lfloor {\sqrt{n}} \right\rfloor}{4n^{2}-1} &=  \frac{6 + {\pi}^2}{24}, \quad p\in\mathbb{R},\,p\gt{1}  \\ \notag \\      \sum\limits_{n=1}^{\infty}\frac{\left \lfloor {{n}^{\frac{1}{p}}} \right\rfloor}{4n^{2}-1} &=  \frac{1}{4} {\left( {1 + \sum\limits_{n=1}^{\infty} \frac{1}{{n}^{p}}}\right)}, \quad p\in\mathbb{R},\,p\gt{1}  \\ \notag \\        \sum\limits_{n=0}^{\infty} x^{\left \lfloor {\sqrt{n}} \right\rfloor} &= \frac{1+x}{(1-x)^2}, \quad x\in\mathbb{R},\, \mid x \mid \lt{1}  \\ \notag \\     \sum\limits_{n=1}^{\infty}\frac{\left \lfloor {\log_p(n)} \right\rfloor}{n(n+1)} &= \frac{1}{p-1}, \quad p\in\mathbb{R},\,p\gt{1}  \\ \notag \\ \sum\limits_{n=1}^{\infty}\frac{\left \lfloor {\log_p(n)} \right\rfloor}{4n^{2}-1} &=  \frac{1}{4\,(p-1)}, \quad p\in\mathbb{R},\,p\gt{1}  \\ \notag \\ \sum\limits_{n=1}^{\infty} x^{\left \lfloor {\log_p(n)} \right\rfloor} &= \frac{p-1}{1-p\,x}, \quad p,x\in\mathbb{R},\,p\gt{1},\,p\,x\lt{1}  \\ \notag \\ \sum\limits_{n=1}^{\infty} \frac{1}{p^{\left \lfloor {\log_{p-1}(n)} \right\rfloor}} &={p}\,{(p-2)}, \quad p\in\mathbb{R},\,p\gt{2}  \\ \notag \\ \sum\limits_{n=1}^{\infty} \frac{1}{p^{\left \lfloor {\log_{2}(n)} \right\rfloor}} &=\frac{p}{p-2}, \quad p\in\mathbb{R},\,p\gt{2}  \\ \notag \\ \cdots &= \cdots  \end{align}  \]

   

Kommentare sind geschlossen.