Write out each sum:
(a) \(\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k}\)
(b) \(\displaystyle\sum\limits_{k=1}^{n}k!\)
Solution (a) \(\displaystyle\sum\limits_{k=1}^{n}\dfrac{1}{k}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}\)
(b) \(\displaystyle\sum\limits_{k=1}^{n}k!=1!+2!+\cdots+n!\)