이중 시그마 합의 값

$\displaystyle \sum\limits _{k = 1} ^{20} ( 2 k + 1 ) \left( \frac{1}{k} + \frac{1}{k + 1} + \frac{1}{k + 2} + \cdots + \frac{1}{20} \right)$ $\displaystyle = \sum\limits _{k = 1} ^{1} ( 2 k + 1 ) \left( \frac{1}{1} \right) + \sum\limits _{k = 1} ^{2} ( 2 k + 1 ) \left( \frac{1}{2} \right) + \sum\limits _{k = 1} ^{3} ( 2 k + 1 ) \left( \frac{1}{3} \right) +$ $\displaystyle \cdots + \sum\limits _{k = 1} ^{20} ( 2 k + 1 ) \left( \frac{1}{20} \right)$ $\displaystyle = \sum\limits _{n = 1} ^{20} n ( n + 2 ) \times \frac{1}{n}$ $\displaystyle = \sum\limits _{n = 1} ^{20} ( n + 2 )$ $= 250$