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) $\displaystyle\sum\limits_{k=1}^{n}k!=1!+2!+\cdots+n!$