Σ1/n!{1/(n+1)-1/(n+2)}=Σ1/(n+2)!を考える

Σ1/n!{1/(n+1)}=Σ1/(n+1)!

Σ1/n!(n+2)=Σ1/(n+1)!-Σ1/(n+2)!={1/2!+1/3!+1/4!+・・・}-{1/3!+1/4!+1/5!+・・・}

=1/2

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Σ1/(n+1)!{1/(n+2)-1/(n+3)}=Σ1/(n+3)!を考える

Σ1/(n+1)!{1/(n+2)}=Σ1/(n+2)!

Σ1/(n+1)!(n+3)=Σ1/(n+2)!-Σ1/(n+3)!={1/3!+1/4!+1/5!+・・・}-{1/4!+1/5!+1/6!+・・・}

=1/6

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