Not counting all the terms

Level 1

lim n r = 1 n 1 ( r + 2 ) r ! \large \displaystyle\lim_{n \to \infty}\sum_{r=1}^{n}\frac{1}{(r+2)r!}

Find the limit of the infinite sum above.

1 9 \frac19 2 1 2 \frac12 1 4 \frac14

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1 solution

Otto Bretscher
May 12, 2015

r = 1 ( 1 ( r + 1 ) ! 1 ( r + 2 ) ! ) = ( e 1 1 ) ( e 1 1 1 2 ! ) = 1 2 \sum_{r=1}^{\infty}\left(\frac{1}{(r+1)!}-\frac{1}{(r+2)!}\right)=(e-1-1)-\left(e-1-1-\frac{1}{2!}\right)=\frac{1}{2}

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