Question

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

Answer 1

`S_6=49/20`

Explanation:

Let's get a formula for the `kth` partial sum:

`S_k=sum_(n=1)^k1/n=1+1/2+1/3+...+1/k`

The `6th` partial sum is then obtained as follows:

`S_6=sum_(n=1)^6 1/n=1+1/2+1/3+1/4+1/5+1/6=49/20`