# How do you find the 6-th partial sum of the infinite series sum_(n=1)^oo1/n ?

Apr 21, 2018

${S}_{6} = \frac{49}{20}$

#### Explanation:

Let's get a formula for the $k t h$ partial sum:

${S}_{k} = {\sum}_{n = 1}^{k} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{k}$

The $6 t h$ partial sum is then obtained as follows:

${S}_{6} = {\sum}_{n = 1}^{6} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{49}{20}$