Where does it go?

Though the numerical value of this sequence tends to grow slowly, it actually diverges to infinity.

To prove this, first consider the sequence S1= $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}...$ . Now rewrite it as S1 = $1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}...$ and apparently this sequence will diverge.

The last step is to compare S1 to our original sequence: Original sequence = $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}...$ S1= $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}...$

Obviously, the original sequence appears to have larger value than S1. Since S1 diverges, this sequence must diverge too.

The Harmonic series, as this series are called, diverges . While Margaret provided simple proof, I will use the integral. For $1＋\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$ This is equal to the area of rectangle of the size $1\times\frac{1}{n}$ . The area below the graph $y = \frac{1}{x}$ from $x = 1$ to $n$ is defined by $\int_{1}^{n}\frac{dx}{x} = ln(n) < 1＋\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{n}$ Here, as $n$ reaches infinity, the value of $ln(n)$ will reach its infinity as well, which, in turn, cause the series to diverge.