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The value of $\sum_{r=2}^{\infty} \frac{1+2+\quad+(r-1)}{r !}$
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Verified Answer
The correct answer is:
$\frac{e}{2}$
$\sum_{r=2}^{\infty} \frac{(r-1) r}{2 r !}=\sum_{r=2}^{\infty} \frac{1}{2(r-2) !}$
$=\frac{1}{2}\left[\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\quad \infty\right]=\frac{1}{2} e$
$=\frac{1}{2}\left[\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\quad \infty\right]=\frac{1}{2} e$
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