Find the sum of the harmonic sequence: 1/12 + 1/24 + 1/36 + 1/48 + 1/60 + ...

The given sequence is $\frac{1}{12}, \frac{1}{24}, \frac{1}{36}, \frac{1}{48}, \frac{1}{60}, \dots$.
Consider the reciprocals of these terms: $12, 24, 36, 48, 60, \dots$.
This sequence of reciprocals is an arithmetic progression with a first term $a_1 = 12$ and a common difference $d = 24 - 12 = 12$.
Therefore, the original sequence is a harmonic progression.

The $n$-th term of the arithmetic progression is $A_n = a_1 + (n-1)d = 12 + (n-1)12 = 12 + 12n - 12 = 12n$.
The $n$-th term of the harmonic progression is $H_n = \frac{1}{A_n} = \frac{1}{12n}$.

The sum of the harmonic sequence is given by the infinite series:

Factor out the constant $\frac{1}{12}$:

The series $\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series. The harmonic series is known to diverge, meaning its sum does not approach a finite value.
Since the harmonic series diverges, multiplying it by a non-zero constant ($\frac{1}{12}$) also results in a divergent series.

The sum of the sequence is divergent.