Why the harmonic series diverges (but $\sum 1/n^2$ doesn't)

The theorem

• $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n}$ diverges. (Harmonic series)
• $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^\alpha}$ converges iff $\alpha > 1$. (Riemann series)

Elegant proof by grouping (Oresme, 1350)

Group terms by sizes $1, 2, 4, 8, \dots$:

$1 + \dfrac{1}{2} + \underbrace{\left(\dfrac{1}{3} + \dfrac{1}{4}\right)}_{> 2 \times 1/4 = 1/2} + \underbrace{\left(\dfrac{1}{5} + \dots + \dfrac{1}{8}\right)}_{> 4 \times 1/8 = 1/2} + \dots$

Each group is strictly more than $1/2$. Since there’s an infinity of groups, the sum is infinite.

Why $\sum 1/n^2$ converges

Decomposition by telescoping: $\dfrac{1}{n^2} < \dfrac{1}{n(n-1)} = \dfrac{1}{n-1} - \dfrac{1}{n}$.

Summing, everything telescopes and the sum is bounded by 2. In fact, $\displaystyle \sum \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$ (Euler, 1735).

The Riemann criterion rule

$\dfrac{1}{n^\alpha}$ converges iff the power exceeds 1. Applications:

• $\sum 1/\sqrt{n} = \sum 1/n^{1/2}$ → diverges ($1/2 < 1$)
• $\sum 1/n^3$ → converges
• $\sum 1/(n \ln n)$ → diverges (integral comparison)

Useful comparisons

• If $|u_n| \leq v_n$ and $\sum v_n$ converges, then $\sum u_n$ converges.
• If $u_n \sim v_n$ at infinity (same signs), $\sum u_n$ and $\sum v_n$ have the same nature.

?Your turn

Which of these series converges?

A.Σ 1/nB.Σ 1/√nC.Σ 1/n^{1.5}D.Σ 1/ln(n)

Check