Another Proof that harmonic series diverges.

Prove $$ \sum\frac{1}{n} $$ diverges by comparing with $$\sum a_n$$ where $a_n$ is the sequence $$(\frac{1}{2}, \frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8},\frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16},\frac{1}{16}, \frac{1}{32}, \frac{1}{32}, ....)$$

Solution 1:

This is one of the standard ways to prove that the harmonic series diverges.

HINT

1. Find the general form of $a_n$.
2. Prove that $$\dfrac1n > a_n$$
3. Find $$\displaystyle \sum_{n=1}^{n=2^m-1} a_n$$
4. From ($2$), we have $$\displaystyle \sum_{n=1}^{n=2^m-1} \dfrac1n > \displaystyle \sum_{n=1}^{n=2^m-1} a_n$$
5. Conclude that $$\displaystyle \sum_{n=1}^{n=\infty} \dfrac1n $$ diverges by letting $m \to \infty$ in ($4$).