Endliche Summen(2)

Für $p \in \mathbb{N}$ :

(1) $\begin{equation*} \sum\limits_{k=0}^{n}{\left\lfloor {\frac{k}{p}} \right\rfloor} = {\left\lfloor {\frac{n}{p}} \right\rfloor} {\left({ n+1\, - \,\frac{p}{2}\left({{\left\lfloor {\frac{n}{p}} \right\rfloor} +1 }\right)}\right)} \end{equation*}$

(2) $\begin{equation*} \sum\limits_{k=0}^{n}{\left\lfloor {\frac{k}{p}} \right\rfloor} = \left(n+1\right)\left\lfloor \frac{n}{p} \right\rfloor - \,p \cdot S_{\left\lfloor \frac{n}{p} \right\rfloor}^{(1)} \end{equation*}$

(3) $\begin{equation*} \begin{split} \sum\limits_{k=0}^{n}{(-1)^k\left\lfloor {\frac{k}{p}} \right\rfloor} &= \frac{1}{8} {\left[ 2 \left(1 + (-1)^p+2(-1)^n \right) {\left\lfloor {\frac{n}{p}} \right\rfloor} \right]} \\ &\quad -\frac{1}{8} {\left[{\left( 1-(-1)^p \right)} {\left( 1-(-1)^{\left\lfloor {\frac{n}{p}} \right\rfloor} \right)} \right]} \\&= \frac{1}{2} (-1)^n {\left\lfloor {\frac{n}{p}} \right\rfloor} - \frac{1}{2} \begin{cases} {-\left\lfloor {\frac{n}{p}} \right\rfloor} &\textrm{if } p\equiv 0 \pmod{2} \\ {\left\lfloor {\frac{n}{p}} \right\rfloor} \bmod{2} &\textrm{if } p\equiv 1 \pmod{2} \end{cases} \end{split} \end{equation*}$

(4) $\begin{align*} \sum\limits_{k=0}^{n}{k\left\lfloor {\frac{k}{p}} \right\rfloor} &= \frac{1}{2}{\left\lfloor {\frac{n}{p}} \right\rfloor} \left[ {{n\,(n+1)}\, - \frac{p}{6} {\left({{\left\lfloor {\frac{n}{p}} \right\rfloor} + 1} \right)} {\left( {{2\, p\left\lfloor {\frac{n}{p}} \right\rfloor} + p\,\, - 3}\right)}}\right] \\ &= \frac{{\left\lfloor {\frac{n}{p}} \right\rfloor} \left[ {{6\,n\,(n+1)}\, - {2\,p^2}{\left\lfloor {\frac{n}{p}} \right\rfloor}^2 } - 3\,p\,(p-1){\left\lfloor {\frac{n}{p}} \right\rfloor} - p\,(p-3)\right]}{12} \end{align*}$

(5) $\begin{equation*} \begin{split} \sum\limits_{k=0}^{n}{(-1)^k k\left\lfloor {\frac{k}{p}} \right\rfloor} &= {\left\lfloor {\frac{n+1}{2}} \right\rfloor} {\left\lfloor {\frac{n}{p}} \right\rfloor} (-1)^n + \sum\limits_{j=1}^{\left\lfloor {\frac{n}{p}} \right\rfloor}{(-1)^{pj} \left\lfloor {\frac{pj}{2}} \right\rfloor} \\ &= \frac{2n+1}{4} {\left\lfloor {\frac{n}{p}} \right\rfloor}(-1)^n + \\ &\quad + \begin{cases} \frac{p}{2} \frac{{\left\lfloor {\frac{n}{p}} \right\rfloor}\left(\left\lfloor {\frac{n}{p}} \right\rfloor + 1\right)}{2} - \frac{1}{4}{\left\lfloor {\frac{n}{p}} \right\rfloor} &\textrm{if } p\equiv 0 \pmod{2} \\ \frac{p}{2} {\left\lfloor \frac{{\left\lfloor {\frac{n}{p}} \right\rfloor} + 1}{2} \right\rfloor} (-1)^{\left\lfloor {\frac{n}{p}} \right\rfloor} + \frac{1}{4} \frac{1 - (-1)^{\left\lfloor {\frac{n}{p}} \right\rfloor}}{2} &\textrm{if } p\equiv 1 \pmod{2} \end{cases} \end{split} \end{equation*}$

(6) $\begin{equation*} {S_{n}}^{(r)}:=\sum\limits_{k=1}^{n}{\left\lfloor {k^{r}} \right\rfloor} , \, r \in \mathbb{R}, \quad r > {0} \end{equation*}$

(7) $\begin{equation*} \begin{split} \sum\limits_{k=1}^{n}{\left\lfloor {\sqrt{k}} \right\rfloor} &= \left( {n+1} \right) \left\lfloor {\sqrt{n}}\, \right\rfloor - S_{\left\lfloor {\sqrt{n}}\, \right\rfloor}^{(2)} \\ &= \left( {n+1} \right) \left\lfloor {\sqrt{n}}\, \right\rfloor - \frac{{\left\lfloor {\sqrt{n}} \right\rfloor \left( {\left\lfloor {\sqrt{n}} \right\rfloor+1} \right) \left( {2\left\lfloor {\sqrt{n}} \right\rfloor+1} \right)}}{6} \end{split} \end{equation*}$

(8) $\begin{equation*} \begin{split} \sum\limits_{k=1}^{n}{\left\lfloor {\sqrt{k}} \right\rfloor}^2 &= \left( {n+1} \right) {\left\lfloor {\sqrt{n}}\, \right\rfloor}^2 + S_{\left\lfloor {\sqrt{n}}\, \right\rfloor}^{(2)} - 2\cdot S_{\left\lfloor {\sqrt{n}}\, \right\rfloor}^{(3)} \\ &= \left( {n+1} \right) {\left\lfloor {\sqrt{n}}\, \right\rfloor}^2 + \frac{{\left\lfloor {\sqrt{n}} \right\rfloor \left( {\left\lfloor {\sqrt{n}} \right\rfloor+1} \right) \left( {2\left\lfloor {\sqrt{n}} \right\rfloor+1} \right)}}{6} \\ &\quad - \frac{{{\left\lfloor {\sqrt{n}} \right\rfloor}^2 {\left( {\left\lfloor {\sqrt{n}} \right\rfloor+1} \right)}}^2}{2} \end{split} \end{equation*}$

(9) $\begin{equation*} \sum\limits_{k=1}^{n}{\left \lfloor {k^\frac{1}{3}} \right\rfloor} = \left( {n+1} \right) {\left \lfloor {n^\frac{1}{3}} \right\rfloor} -\frac{{\left \lfloor {n^\frac{1}{3}} \right\rfloor}^{2}\,{\left( {{\left \lfloor {n^\frac{1}{3}} \right\rfloor} +1} \right)}^2}{4}\end{equation*}$

Für $p\in\mathbb{R},\,p \ge {1}$ :

(10) $\begin{equation*} \sum\limits_{k=1}^{n}{\left \lfloor {k^\frac{1}{p}} \right\rfloor}=\left( {n+1} \right)\cdot {\left\lfloor {n^\frac{1}{p}} \right\rfloor} - \sum\limits_{k=1}^{\left\lfloor {n^\frac{1}{p}} \right\rfloor}{{k}^{p}} \end{equation*}$

(11) $\begin{equation*} S_{n}^{\left(\frac{1}{p}\right)}+{S}_{\left\lfloor {n^\frac{1}{p}} \right\rfloor}^{(p)} = \left( {n+1} \right)\cdot \left \lfloor {n^\frac{1}{p}} \right\rfloor \end{equation*}$

(12) $\begin{equation*} \sum\limits_{k=1}^{n}{k\left\lfloor {{k^{\frac{1}{p}}}} \right\rfloor} = \frac{1}{2} \left( { n(n+1) {\left\lfloor {n^{\frac{1}{p}}} \right\rfloor} + S_{\left\lfloor {n^{\frac{1}{p}}} \right\rfloor}^{(p)} - S_{\left\lfloor {n^{\frac{1}{p}}}\right\rfloor}^{(2p)} }\right) \end{equation*}$

(13) $\begin{equation*} \begin{split} \sum\limits_{k=1}^{n}{(-1)^k\,\left \lfloor {k^\frac{1}{p}} \right\rfloor} &= \frac{1}{4} \left( 2 \,(-1)^n {\left\lfloor {n^\frac{1}{p}} \right\rfloor} + {(-1)^{\left\lfloor {n^\frac{1}{p}} \right\rfloor} - 1} \right) \\ &= \frac{1}{2} \left( (-1)^n {\left\lfloor {n^\frac{1}{p}} \right\rfloor} + {(-1)^{\left\lfloor {n^\frac{1}{p}} \right\rfloor}}\bmod {2} \right) \end{split} \end{equation*}$

(14) $\begin{equation*}\sum\limits_{k=1}^{n}{\left\lfloor {\log_2{(k)}} \right\rfloor} = ({n+1}) \left\lfloor {\log_2{(n)}} \right\rfloor - 2 \left({2^{ \left\lfloor {\log_2{(n)}} \right\rfloor} -1} \right)\end{equation*}$

(15) $\begin{equation*}\sum\limits_{k=1}^{n}{\left\lfloor {\log{(k)}} \right\rfloor} = ({n+1}) \left\lfloor {\log{(n)}} \right\rfloor - \frac{10}{9} \left({10^{ \left\lfloor {\log{(n)}} \right\rfloor} -1} \right)\end{equation*}$

Für $p\in \mathbb{N}, \, p > 1$ :

(16) $\begin{equation*} \sum\limits_{{k=1}}^{n}{\left\lfloor {\log_p{(k)}} \right\rfloor} = ({n+1}) \left\lfloor {\log_p{(n)}} \right\rfloor - \frac{p}{p-1} \left({p^{ \left\lfloor {\log_p{(n)}} \right\rfloor} -1} \right) \end{equation*}$

(17) $\begin{equation*} \sum\limits_{{k=1}}^{n}{k\,\left\lfloor {\log_p{(k)}} \right\rfloor} = \frac{n(n+1)}{2} \left\lfloor {\log_p{(n)}} \right\rfloor - \frac{p}{2} \frac{\left({p^{ \left\lfloor {\log_p{(n)}} \right\rfloor} -1} \right) \left({p^{ \left\lfloor {\log_p{(n)}} \right\rfloor + 1} -1} \right)}{p^2-1} \end{equation*}$

(18) $\begin{equation*} \sum\limits_{{k=1}}^{n}{(-1)^k\,\left\lfloor {\log_p{(k)}} \right\rfloor} &= \frac{(-1)^p+(-1)^n}{2} \left\lfloor {\log_p{(n)}} \right\rfloor \end{equation*}$

Endliche Summen(2)

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