INTEGRIAS PROGRESSIMAIS SEQUENCIAIS INFINITESIMAIS DE GRACELI EM SOMAS.

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${\displaystyle {\overline {X}}={\frac {1}{n}}\sum _{i=1}^{n}x_{i},}$  k[pr]ph] DX =  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m}^{n}\alpha x_{i}=\alpha \sum _{i=m}^{n}x_{i}}$^[3]  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m}^{n}(x_{i}\pm y_{i})=\sum _{i=m}^{n}x_{i}\pm \sum _{i=m}^{n}y_{i}}$^[3]  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m}^{m}x_{i}=x_{m}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m}^{n}x_{i}=\sum _{i=m}^{p}x_{i}+\sum _{i=p+1}^{n}x_{i},\quad \forall m\leq p\leq n}$^[3]  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m}^{n}x_{i}=\sum _{i=m+p}^{n+p}x_{i-p}}$^[3]  k[pr]ph] DX =

  k[pr]ph] DX = ${\displaystyle \sum _{i=m}^{n}(x_{i+1}-x_{i})=x_{n+1}-x_{m}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m}^{n}\sum _{j=k}^{l}x_{i}y_{j}=\sum _{i=m}^{n}x_{i}\sum _{j=k}^{l}y_{j}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \left|\sum _{i=m}^{n}x_{i}\right|\leq \sum _{i=m}^{n}|x_{i}|}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{n=0}^{t}x_{2n}+\sum _{n=0}^{t}x_{2n+1}=\sum _{n=0}^{2t+1}x_{n}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{n=0}^{t}\sum _{i=0}^{z-1}x_{z\cdot n+i}=\sum _{n=0}^{z\cdot t+z-1}x_{n}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \left(\sum _{k=0}^{n}a_{k}\right)\cdot \left(\sum _{k=0}^{n}b_{k}\right)=\sum _{k=0}^{2n}\sum _{i=0}^{k}a_{i}b_{k-i}-\sum _{k=0}^{n-1}\left(a_{k}\sum _{i=n+1}^{2n-k}b_{i}+b_{k}\sum _{i=n+1}^{2n-k}a_{i}\right)}$  k[pr]ph] DX =

$${\displaystyle \sum _{i,j=1}^{n}:=\sum _{i=1}^{n}\sum _{j=1}^{n}x_{i,j}}$$

  k[pr]ph] DX   k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{k\leq j\leq i\leq n}x_{i,j}:=\sum _{i=k}^{n}\sum _{j=k}^{i}x_{i,j}=\sum _{j=k}^{n}\sum _{i=j}^{n}x_{i,j}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{i=m+k}^{n}\sum _{j=m}^{i-k}a_{i,j}=\sum _{j=m}^{n-k}\sum _{i=j+k}^{n}a_{i,j}}$  k[pr]ph] DX =

$${\displaystyle \prod _{i=1}^{n}x_{i}:=x_{1}x_{2}x_{3}\cdots x_{n}}$$

  k[pr]ph] DX =  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle \sum _{n=s}^{t}\ln x_{n}=\ln \prod _{n=s}^{t}x_{n}}$  k[pr]ph] DX =

  k[pr]ph] DX =. ${\displaystyle c^{\left[\sum _{n=s}^{t}x_{n}\right]}=\prod _{n=s}^{t}c^{x_{n}},{\mbox{onde}}~c>0.}$  k[pr]ph] DX =

$${\displaystyle \sum _{i=m}^{n}{x_{i}}}$$

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1) ${\displaystyle \sum _{i=1}^{5}2i}$

$${\displaystyle t=n+1-m-r=5+1-1-0}$$

$${\displaystyle \sum _{i=1}^{5}2i=2(1)+2(2)+2(3)+2(4)+2(5)=2+4+6+8+10}$$

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1. ${\displaystyle \sum _{i=1}^{n}i^{3}=1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}}$^[3]
2. ${\displaystyle \sum _{i=1}^{n}i^{4}=1^{4}+2^{4}+3^{4}+\cdots +n^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}}$  k[pr]ph] DX =^[3]  k[pr]ph] DX =
3. ${\displaystyle \sum _{i=1}^{n}i^{5}=1^{5}+2^{5}+3^{5}+\cdots +n^{5}={\frac {n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}}}$  k[pr]ph] DX =^[3]  k[pr]ph] DX =
4. ${\displaystyle \sum _{i=1}^{n}i^{6}=1^{6}+2^{6}+3^{6}+\cdots +n^{6}={\frac {n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}}}$  k[pr]ph] DX =^[3]  k[pr]ph] DX =
5. ${\displaystyle \sum _{i=0}^{n}i^{p}={\frac {(n+1)^{p+1}}{p+1}}+\sum _{k=1}^{p}{\frac {B_{k}}{p-k+1}}{p \choose k}(n+1)^{p-k+1}}$  k[pr]ph] DX =  k[pr]ph] DX =

1. ${\displaystyle \sum _{i=m}^{n}x^{i}=x^{m}+x^{m+1}+x^{m+2}+\cdots +x^{n}={\frac {x(x^{n}-x^{m-1})}{x-1}}}$  k[pr]ph] DX =   k[pr]
2. ${\displaystyle \sum _{i=0}^{n-1}ia^{i}={\frac {a-na^{n}+(n-1)a^{n+1}}{(1-a)^{2}}},}$  k[pr]ph] DX = ${\displaystyle (a\neq 1)}$^[3]  k[pr]ph] DX =

$${\displaystyle \sum _{p=1}^{n}{\frac {1}{p}}=1+{\frac {1}{2}}+\cdots }$$

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$${\displaystyle \sum _{p=1}^{n}{\frac {1}{p+1}}={\frac {1}{2}}+{\frac {1}{3}}.}$$

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Para ${\displaystyle n=3:}$

$${\displaystyle \sum _{p=1}^{n}{\frac {1}{p}}=1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots }$$

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$${\displaystyle \sum _{p=1}^{n}{\frac {1}{p+1}}={\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots .}$$

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$${\displaystyle \sum _{p=1}^{n}{\frac {1}{p}}-\sum _{p=1}^{n}{\frac {1}{p+1}}=1-{\frac {1}{n+1}}}$$

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$${\displaystyle \sum _{p=1}^{n}\left({\frac {1}{p}}-{\frac {1}{p+1}}\right)=\sum _{p=1}^{n}{\frac {1}{p(p+1)}}={\frac {n}{n+1}}.}$$

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Logo:

$${\displaystyle \sum _{p=1}^{n}{\frac {1}{p(p+1)}}={\frac {n}{n+1}}.}$$

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