[Figure: FiveStudy logo]
FiveStudy
Calculus BC · Unit 10 · 10.5
AP Calculus BC · Infinite Sequences and Series
Harmonic Series and p-Series
The ruler every other test measures against — and the boundary at p = 1 is sharp.
key idea$\sum_{n=1}^{\infty}\frac{1}{n^{p}}\ \text{converges}\iff p\gt 1$
The test
Applies to
$\displaystyle\sum\frac{1}{n^{p}}$ — a constant power of $n$ in the denominator
Converges when
$p\gt 1$, strictly
Diverges when
$p\le1$, including the harmonic case $p=1$
Harmonic series
$\displaystyle\sum\frac1n$ — the boundary, and it diverges
Never gives
the sum. $\displaystyle\sum\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$, which no AP method produces
The $p$-series family is the ruler the rest of the unit is measured against. Knowing instantly where the boundary sits is what makes comparison tests fast, so this criterion is worth memorising rather than re-deriving.
| Series | $p$ | Verdict |
|---|
| $\displaystyle\sum\frac{1}{n^{3}}$ | $3$ | converges |
| $\displaystyle\sum\frac{1}{n^{2}}$ | $2$ | converges |
| $\displaystyle\sum\frac{1}{n^{1.001}}$ | $1.001$ | converges — barely |
| $\displaystyle\sum\frac{1}{n}$ | $1$ | diverges — harmonic |
| $\displaystyle\sum\frac{1}{\sqrt{n}}$ | $\tfrac12$ | diverges |
| $\displaystyle\sum\frac{1}{\sqrt[3]{n^{2}}}$ | $\tfrac23$ | diverges |
Rows 3 and 4 sit a thousandth apart and land on opposite sides. The boundary at $p=1$ is sharp, and the harmonic series belongs on the divergent side of it.
| Written as | Rewrite as | $p$ |
|---|
| $\dfrac{1}{\sqrt{n}}$ | $n^{-1/2}$ | $\tfrac12$ |
| $\dfrac{1}{n\sqrt{n}}$ | $n^{-3/2}$ | $\tfrac32$ |
| $\dfrac{1}{\sqrt[3]{n}}$ | $n^{-1/3}$ | $\tfrac13$ |
| $\dfrac{n}{n^{3}}$ | $n^{-2}$ | $2$ |
| $\dfrac{1}{2n}$ | $\tfrac12\cdot n^{-1}$ | $1$ — a constant multiple changes nothing |
Getting $p$ right is an exponent-algebra exercise. Convert every root to a fractional power first, and remember that a constant factor out front never affects the verdict.
When to reach for it
| Situation | Use $p$-series… |
|---|
| the term is exactly $n^{-p}$ | directly — read $p$ and decide |
| the term is a ratio of polynomials | as the comparison partner, with $p=$ (denominator degree $-$ numerator degree) |
| the term has $\ln n$ in it | not at all — use the integral test |
| the term alternates | on $\lvert a_n\rvert$ only, to test absolute convergence |
Row 2 is how this test earns its keep. $\dfrac{n^{2}+1}{n^{5}-3}$ behaves like $\dfrac{1}{n^{3}}$, and the difference of the degrees gives $p$ immediately.
Worked examples
EX 1
Read $p$ off a root
convert to a powerDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n\sqrt{n}}$ converges.
1
Rewrite the term
$n\sqrt{n}=n\cdot n^{1/2}=n^{3/2}$.
2
Identify $p$
$p=\dfrac32$.
3
Compare with $1$
$\dfrac32\gt 1$.
4
Conclude
The series converges by the $p$-series criterion.
5
Note what is not claimed
Its sum is not produced by this argument, and no AP technique produces it.
EX 2
The harmonic boundary
$p=1$ divergesDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{5}{3n}$ converges.
1
Pull out the constant
$\dfrac53\displaystyle\sum\dfrac1n$.
2
Identify the family
The remaining series is the harmonic series, with $p=1$.
3
Apply the criterion
$p=1$ is not greater than $1$, so it diverges.
4
Handle the constant
A non-zero constant multiple of a divergent series still diverges.
5
Conclude
The series diverges. Constants can never rescue a divergent series or spoil a convergent one.
EX 3
Degrees decide the partner
comparisonDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{n^{2}+1}{n^{5}-3}$ converges.
1
Find the dominant behaviour
For large $n$ the term behaves like $\dfrac{n^{2}}{n^{5}}=\dfrac{1}{n^{3}}$.
2
Identify the comparison series
A $p$-series with $p=3$, which converges.
3
Apply the limit comparison test
$\displaystyle\lim_{n\to\infty}\dfrac{\left(n^{2}+1\right)/\left(n^{5}-3\right)}{1/n^{3}}=\lim\dfrac{n^{5}+n^{3}}{n^{5}-3}=1$.
4
Read the conclusion
The limit is finite and positive, so both series share a verdict.
5
Conclude
The series converges. The exponent $p=3$ came straight from $5-2$, the difference of the degrees.
EX 4
A near miss
$p$ just below oneDetermine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{0.99}}$ converges.
2
Compare with $1$
$0.99\lt 1$.
3
Conclude
The series diverges.
4
Confront the intuition
Its terms shrink almost as fast as those of the convergent $\dfrac{1}{n^{1.01}}$, yet the verdicts differ.
5
State the moral
The criterion is strict. ‘Close to $1$’ is not a mathematical category; only $p\gt 1$ or $p\le1$ matters.
What it cannot tell you
$p=1$ diverges
The harmonic series is the boundary case and it is on the divergent side. The inequality $p\gt 1$ is strict.
It gives no sums
Convergence is all this criterion produces. $\sum\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6}$ is a famous result well outside the course.
The exponent must be constant
$\sum\dfrac{1}{n^{n}}$ is not a $p$-series. Neither is anything with $\ln n$ in it.
Constant multiples are irrelevant
$\sum\dfrac{7}{n^{2}}$ converges and $\sum\dfrac{1}{100n}$ diverges. Only the exponent decides.