Harmonic series and the p-series threshold

Understand the idea

A p-series sums reciprocal powers of the positive integers. Its convergence depends on whether the exponent is strictly greater than one. The boundary case is the harmonic series, whose terms approach zero but whose partial sums do not approach a finite number.

Rewrite roots and powers into a single exponent before classifying the series. For example, $\displaystyle 1/\sqrt{n^3}=n^{-3/2}$ has $\displaystyle p=3/2$ , not $\displaystyle p=3$ . Multiplying every term by a fixed nonzero constant does not change convergence.

The threshold follows from the integral of $\displaystyle x^{-p}$ over an unbounded interval. At $\displaystyle p=1$ , the antiderivative becomes a logarithm and grows without bound. Grouping harmonic terms into successively larger blocks also reveals why their total cannot settle.

\sum_{n=1}^{\infty}\frac1{n^p}\text{ converges iff }p>1;\qquad \sum_{n=1}^{\infty}\frac1n\text{ diverges}

A method you can apply

Rewrite the term as a constant multiple of $\displaystyle 1/n^p$ when possible.
Identify the exact exponent and compare it with $\displaystyle 1$ .
If another factor is present, check whether the expression is truly a p-series or needs a comparison test.

Worked example

Classify $\displaystyle \sum n^{-5/4}$ and $\displaystyle \sum n^{-3/4}$ , both starting at $\displaystyle n=1$ .

The first is a p-series with $\displaystyle p=5/4>1$ .
The second is a p-series with $\displaystyle p=3/4<1$ .
The terms of both series approach zero, so a zero term limit alone would not distinguish them.

The

\displaystyle 5/4

series converges, while the

\displaystyle 3/4

series diverges.

Try it yourself

Classify $\displaystyle \sum_{n=1}^{\infty}1/\sqrt{n^3}$ and $\displaystyle \sum_{n=1}^{\infty}1/\sqrt[3]{n^2}$ .

Show solution and reasoning

The exponents are $\displaystyle p=3/2$ and $\displaystyle p=2/3$ , respectively. The first converges and the second diverges.