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 Incoming links: Polynomial-logarithmic series convergence

The comparison test for series says that an eventual upper bound by a convergent positive series proves convergence, while an eventual lower bound by a divergent positive series proves divergence. The p-series

\sum n^{- s}

converges exactly when

s > 1

p > 1

, then

(lo g n)^{q} \geq 1

for all sufficiently large

n

, so

0 < \frac{1}{n ^{p} ( l o g n ) ^{q}} \leq \frac{1}{n ^{p}},

and the series converges. If

0 < p < 1

, choose

η = (1 - p) /2 > 0

. The allowed logarithmic bound, applied with exponent

η / q

, gives

(lo g n)^{q} < n^{η}

eventually. Thus

\frac{1}{n ^{p} ( l o g n ) ^{q}} > \frac{1}{n ^{p + η}}, p + η = (p + 1) /2 < 1,

and the series diverges.

For

p = 1

, use the integral test for convergence: for an eventually positive decreasing function

g

\sum g (n)

and

\int g (x) d x

either both converge or both diverge. Here

g (x) = 1/ [x (lo g x)^{q}]

is decreasing for

x > 1

, and the substitution

u = lo g x

gives

\int_{2}^{\infty} \frac{d x}{x ( l o g x ) ^{q}} = \int_{l o g 2}^{\infty} u^{- q} d u .

This integral converges exactly when

q > 1

Convergence iff p > 1 or (p = 1 and q > 1) .

All other allowed positive

p, q