Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 41

$15$

Here, we have $\sum_{i=1}^5 \dfrac{i!}{(i-1)!}= \dfrac{1!}{(1-1)!}+ \dfrac{2!}{(2-1)!}+ \dfrac{3!}{(3-1)!}+ \dfrac{4!}{(4-1)!}+ \dfrac{5!}{(5-1)!}$ $=\dfrac{1}{1}+\dfrac{2}{1}+\dfrac{6}{2}+\dfrac{24}{6}+\dfrac{120}{24}$ $=1+2+3+4+5$ $=15$