数学の基礎と公式-11

数学の基礎と公式

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10. 級数-2

(1)$\displaystyle{}$

$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$より

$\displaystyle{} \sum_{k=1}^n \dfrac{1}{n(n+1)}=1-\dfrac{1}{n+1}$

よって、$n\rightarrow \infty$のとき、

$\displaystyle{ \sum_{k=1}^{\infty} \dfrac{1}{n(n+1)}=1}$

$\displaystyle{ \sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)}={{n\,\left(n+3\right)}\over{4\,\left(n+1\right)\,\left(n+2\right)}}}$より

$\displaystyle{ \sum_{k=1}^{\infty} \dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{4}}$

$\displaystyle{ \sum_{k=1}^{n} \dfrac{1}{k(k+1)(k+2)(k+3)}={{n\,\left(n^2+6\,n+11\right)}\over{18\,\left(n+1\right)\,\left(n+2 \right)\,\left(n+3\right)}}}$より

$\displaystyle{ \sum_{k=1}^{\infty} \dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{18}}$

(2)

$\displaystyle{ \sum_{k=1}^{n} \dfrac{2k+1}{k^2 (k+1)^2}={{n\,\left(n+2\right)}\over{\left(n+1\right)^2}}}$より

$\displaystyle{ \sum_{k=1}^{\infty}\dfrac{2k+1}{k^2 (k+1)^2}=1}$

$\displaystyle{ \sum_{k=1}^{n} \dfrac{3 k^2  + 3 k + 1 }{k^3 (k+1)^3}={{n\,\left(n^2+3\,n+3\right)}\over{\left(n+1\right)^3}}}$より

$\displaystyle{ \sum_{k=1}^{\infty}\dfrac{3 k^2  + 3 k + 1 }{k^3 (k+1)^3}=1}$

$\displaystyle{ \sum_{k=1}^{n} \dfrac{4 k^3  + 6 k^2  + 4 k + 1 }{k^4 (k+1)^4}={{n\,\left(n+2\right)\,\left(n^2+2\,n+2\right)}\over{\left(n+1 \right)^4}}}$より

$\displaystyle{ \sum_{k=1}^{\infty}\dfrac{4 k^3  + 6 k^2  + 4 k + 1 }{k^4 (k+1)^4}=1}$

$\displaystyle{ \sum_{k=1}^{n} \dfrac{1}{(3k-2)(3k+1)(3k+4)}={{n\,\left(3\,n+5\right)}\over{8\,\left(3\,n+1\right)\,\left(3\,n+4 \right)}}}$より

$\displaystyle{ \sum_{k=1}^{\infty}\dfrac{1}{(3k-2)(3k+1)(3k+4)}=\dfrac{1}{24}}$

$\displaystyle{ \sum_{k=0}^{n} \dfrac{1}{(3k+1)(3k+10)}={{\left(n+1\right)\,\left(117\,n^2+618\,n+784\right)}\over{28\, \left(3\,n+4\right)\,\left(3\,n+7\right)\,\left(3\,n+10\right)}}}$より

$\displaystyle{ \sum_{k=0}^{\infty}\dfrac{1}{(3k+1)(3k+10)}=\dfrac{13}{84}}$

$\displaystyle{ \sum_{k=1}^{n} \dfrac{2k+1}{(k^2+1)((k+1)^2+1)}={{n\,\left(n+2\right)}\over{2\,\left(n^2+2\,n+2\right)}}}$より

$\displaystyle{ \sum_{k=1}^{\infty}\dfrac{2k+1}{(k^2+1)((k+1)^2+1)}=\dfrac{1}{2}}$

$\displaystyle{ \sum_{k=0}^{n} \dfrac{2k+1}{(k^2+1)((k+1)^2+1)}={{\left(n+1\right)^2}\over{n^2+2\,n+2}}}$より

$\displaystyle{ \sum_{k=0}^{\infty}\dfrac{2k+1}{(k^2+1)((k+1)^2+1)}=1}$