1に収束する級数一覧

ここでは1に収束（広義）する級数を結果のみ只だ只管列挙する。

$\sum _{n=0}^{\infty }-2=-2-2-2-2-2-\cdots =1$

$\sum _{n=0}^{\infty }(-1)^{n}2=2-2+2-2+2-\cdots =1$

$\sum _{n=0}^{\infty }-2^{n}=-1-2-4-8-16-\cdots =1$

$\sum _{n=0}^{\infty }2^{-n-1}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+{\frac {1}{32}}+\cdots =1$

構文解析に失敗 (SVG（ブラウザのプラグインで MathML を有効にすることができます）: サーバー「https://ja.wikipedia.org/api/rest_v1/」から無効な応答 ("Math extension cannot connect to Restbase."):): {\displaystyle \sum_{n=0}^\infty (-2)^{-n-1}3 =\frac32-\frac{3}{4}+\frac{3}{8} -\frac{3}{16} +\frac{3}{32}-\cdots =1}

$\sum _{n=0}^{\infty }(-2)^{n}3=3-6+12-24+48-\cdots =1$

$\sum _{n=0}^{\infty }2^{-2n-2}3={\frac {3}{4}}+{\frac {3}{16}}+{\frac {3}{64}}+{\frac {3}{256}}+{\frac {3}{1024}}+\cdots =1$

$\sum _{n=0}^{\infty }-2^{2n}3=3+12+{48}+{192}+{768}+\cdots =1$

$\sum _{n=0}^{\infty }-^{n}2^{-2n-2}5={\frac {5}{4}}-{\frac {5}{16}}+{\frac {5}{64}}-{\frac {5}{256}}+{\frac {5}{1024}}-\cdots =1$

$\sum _{n=0}^{\infty }-^{n}2^{2n}5=5-20+{80}-{320}+{1280}-\cdots =1$

$\sum _{n=0}^{\infty }-12n=0-12-24-36-48-\dotsb =1$

$\sum _{n=0}^{\infty }-(-1)^{n}4n=0+4-8+12-16+\dotsb =1$

$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n\ln 2}}={\frac {1}{\ln 2}}-{\frac {1}{2\ln 2}}+{\frac {1}{3\ln 2}}-{\frac {1}{4\ln 2}}+{\frac {1}{5\ln 2}}-\cdots =1$

$\sum _{n=0}^{\infty }{\frac {4}{\pi }}{\frac {(-1)^{n}}{2n+1}}={\frac {4}{\pi }}-{\frac {4}{3\pi }}+{\frac {4}{5\pi }}-{\frac {4}{7\pi }}+{\frac {4}{9\pi }}-\cdots =1$

$\sum _{n=0}^{\infty }{\frac {2e}{e^{2}-1}}{\frac {1}{(2n+1)!}}={\frac {2e}{e^{2}-1}}+{\frac {2e}{3!(e^{2}-1)}}+{\frac {2e}{5!(e^{2}-1)}}+{\frac {2e}{7!(e^{2}-1)}}+{\frac {2e}{9!(e^{2}-1)}}+\cdots =1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!\sin 1}}={\frac {1}{\sin 1}}-{\frac {1}{3!\sin 1}}+{\frac {1}{5!\sin 1}}-{\frac {1}{7!\sin 1}}+{\frac {1}{9!\sin 1}}-\cdots =1$

$\sum _{n=0}^{\infty }{\frac {2e}{e^{2}+1}}{\frac {1}{(2n)!}}={\frac {2e}{e^{2}+1}}+{\frac {2e}{2!(e^{2}+1)}}+{\frac {2e}{4!(e^{2}+1)}}+{\frac {2e}{6!(e^{2}+1)}}+{\frac {2e}{8!(e^{2}+1)}}+\cdots =1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!\cos 1}}={\frac {1}{\cos 1}}-{\frac {1}{2!\cos 1}}+{\frac {1}{4!\cos 1}}-{\frac {1}{6!\cos 1}}+{\frac {1}{8!\cos 1}}-\cdots =1$

$\sum _{n=0}^{\infty }{\frac {1}{n!e}}={\frac {1}{e}}+{\frac {1}{e}}+{\frac {1}{2e}}+{\frac {1}{6e}}+{\frac {1}{24e}}+{\frac {1}{120e}}+\cdots =1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}e=e-e+{\frac {e}{2}}-{\frac {e}{6}}+{\frac {e}{24}}-{\frac {e}{120}}+\cdots =1$

$\sum _{n=0}^{\infty }-{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!}}=-1+{\frac {\pi ^{3}}{3!}}-{\frac {\pi ^{5}}{5!}}+{\frac {\pi ^{7}}{7!}}-{\frac {\pi ^{9}}{9!}}+\cdots =1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n}}{(2n)!2^{2n}}}=1-{\frac {\pi ^{2}}{2!2^{2}}}+{\frac {\pi ^{4}}{4!2^{4}}}-{\frac {\pi ^{6}}{6!2^{6}}}+{\frac {\pi ^{8}}{8!2^{8}}}-\cdots =1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n+1}}{(2n+1)!2^{4n+1.5}}}=1-{\frac {\pi ^{3}}{3!2^{5.5}}}+{\frac {\pi ^{5}}{5!2^{9.5}}}-{\frac {\pi ^{7}}{7!2^{13.5}}}+{\frac {\pi ^{9}}{9!2^{17.5}}}-\cdots =1$

$\sum _{n=0}^{\infty }{\frac {(-1)^{n}\pi ^{2n}}{(2n)!2^{4n-{\frac {1}{2}}}}}=1-{\frac {\pi ^{2}}{2!2^{3.5}}}+{\frac {\pi ^{4}}{4!2^{7.5}}}-{\frac {\pi ^{6}}{6!2^{11.5}}}+{\frac {\pi ^{8}}{8!2^{15.5}}}-\cdots =1$

$\sum _{n=0}^{\infty }-{\frac {(i\pi )^{n}}{n!}}=-1-i\pi +{\frac {\pi ^{2}}{2}}+{\frac {i\pi ^{3}}{6}}-{\frac {\pi ^{4}}{24}}-{\frac {i\pi ^{5}}{120}}+\cdots =1$

$\sum _{n=0}^{\infty }{\frac {2i^{n}}{1+i}}={\frac {2}{1+i}}+{\frac {2i}{1+i}}-{\frac {2}{1+i}}-{\frac {2i}{1+i}}+{\frac {2}{1+i}}+\cdots =1$

$\sum _{n=0}^{\infty }{\frac {2i^{-n}}{1-i}}={\frac {2}{1-i}}-{\frac {2i}{1-i}}-{\frac {2}{1-i}}+{\frac {2i}{1-i}}+{\frac {2}{1-i}}-\cdots =1$

独自研究

このページには独自研究が含まれている可能性があります。 エンペディアでは、噂や推測についても有用性があれば掲載可能です。 ぜひ、独自研究の増加にご協力ください。