Iterated Logarithmic Thresholds

Why Iterated Logarithms Create Successive Thresholds

For a series with terms involving $n$ and $\ln n$, the exponent on $\ln n$ can decide convergence when the factor $1/n$ is exactly at its harmonic threshold. If that exponent is itself critical, another logarithm becomes decisive. This process can continue through several layers.

The resulting rule is a sequence of thresholds: examine the exponent on $\ln n$; if it equals one, examine the exponent on $\ln\ln n$; if that also equals one, move to the next iterated logarithm. At the first exponent that is not one, its position relative to one determines the outcome. The last logarithmic factor has the familiar threshold of a power integral.

Notation and the Threshold Rule

Define the iterated logarithms recursively, wherever the expressions are defined, by $L_1(x)=\ln x$ and $L_{j+1}(x)=\ln(L_j(x))$. For any fixed finite number of iterations, all the $L_j(x)$ are positive once $x$ is sufficiently large. A finite number of initial terms does not affect whether a series converges, so all series below are understood to start far enough out that their terms are positive and defined.

Definition: For fixed real exponents $p_1,\ldots,p_m$, consider the positive series with general term $$ a_n=\frac{1}{n\,L_1(n)^{p_1}L_2(n)^{p_2}\cdots L_m(n)^{p_m}}. $$ The exponent $p_j$ is critical at the value $1$ when all the preceding exponents $p_1,\ldots,p_{j-1}$ equal $1$.

The classification uses two standard facts. First, at a stage where the current exponent is different from one, fixed powers of the later logarithmic factors do not change the convergence decision made by that exponent. This is the away-from-critical-power comparison established in the earlier material on logarithmic factors away from the critical power. Second, when the current exponent is exactly one, a logarithmic substitution removes that factor and exposes the next one.

Theorem: For the series above, find the first index $j$ for which $p_j\ne 1$, if one exists. If $p_j>1$, the series converges; if $p_j<1$, it diverges. If every exponent equals $1$, the series diverges. Equivalently, if $p_1=\cdots=p_{m-1}=1$, the final factor gives convergence exactly when $p_m>1$.

Thus “first exponent different from one” includes the final exponent. If it is the final exponent and exceeds one, the series converges; if it is one or less, the series diverges. The rule concerns a finite product of iterated logarithms with fixed real exponents.

Why the Successive Substitutions Work

The integral test compares a positive, eventually decreasing function with its corresponding series. For the function associated with $a_n$, eventual decrease holds: its logarithmic derivative is

\frac{d}{dx}\ln\!\left( \frac{1}{xL_1(x)^{p_1}\cdots L_m(x)^{p_m}} \right) = -\frac{1}{x}\left( 1+\frac{p_1}{L_1(x)} +\frac{p_2}{L_1(x)L_2(x)} +\cdots+ \frac{p_m}{L_1(x)\cdots L_m(x)} \right).

As $x$ grows, every fraction inside the parentheses tends to zero, so the parentheses tend to $1$. The derivative is therefore negative for all sufficiently large $x$. The integral test applies after discarding any finite initial segment.

Suppose first that $p_1=\cdots=p_{j-1}=1$. In the integral, substitute successively $u_1=\ln x$, then $u_2=\ln u_1$, and so on. Each substitution uses $du_k=du_{k-1}/u_{k-1}$. After the first $j-1$ critical factors have been removed, the remaining integral has the form

\int \frac{du}{u^{p_j}(\ln u)^{p_{j+1}}\cdots},

where $u$ represents the $j$-th iterated logarithm. If $p_j\ne1$, the away-from-critical-power comparison decides the outcome: the factor $u^{-p_j}$ is on the convergent side for $p_j>1$ and on the divergent side for $p_j<1$, regardless of the fixed later logarithmic powers.

If the last exponent is reached after all earlier exponents equal one, the substitutions reduce the question to an ordinary power integral:

\int^\infty \frac{du}{u^{p_m}}.

This integral converges for $p_m>1$ and diverges for $p_m\le 1$. In particular, equality at the final threshold is not convergence: $\int du/u$ has antiderivative $\ln u$, which grows without bound.

Worked Examples at Different Levels

Worked Example: A Convergent Series at the Second Logarithm

Classify

\sum_{n} \frac{1}{n\ln n\,(\ln\ln n)^{3/2}},

where the sum starts sufficiently far out that both logarithms are positive. Here $p_1=1$, so the first logarithm is critical; the next exponent is $p_2=3/2>1$. The threshold rule predicts convergence. To see the decisive integral directly, use the integral test and substitute $u=\ln x$, so $du=dx/x$:

\int^\infty \frac{dx}{x\ln x\,(\ln\ln x)^{3/2}} = \int^\infty \frac{du}{u(\ln u)^{3/2}}.

Now put $v=\ln u$, so $dv=du/u$. Then

\int^\infty \frac{du}{u(\ln u)^{3/2}} = \int^\infty v^{-3/2}\,dv = \left[-2v^{-1/2}\right]_{v_0}^{\infty} = \frac{2}{\sqrt{v_0}},

for any starting value $v_0>0$. The limit at infinity is zero because $v^{-1/2}\to0$. The improper integral is finite, so the series converges. As a check, differentiating the antiderivative gives $\frac{d}{dv}(-2v^{-1/2})=v^{-3/2}$, the integrand after substitution.

Worked Example: The Final Critical Exponent Diverges

Consider the series

\sum_n \frac{1}{n\,L_1(n)L_2(n)L_3(n)} = \sum_n \frac{1}{n\ln n\,\ln\ln n\,\ln\ln\ln n}.

All three exponents equal one. The theorem therefore predicts divergence. The substitutions show why the last equality still diverges. Start with the corresponding integral, with a lower limit large enough that every iterated logarithm is positive. Set $u=\ln x$, so $du=dx/x$:

\int \frac{dx}{xL_1(x)L_2(x)L_3(x)} = \int \frac{du}{u(\ln u)(\ln\ln u)}.

Next set $w=\ln u$, so $dw=du/u$. Since $\ln u=w$ and $\ln\ln u=\ln w$, the result is

\int \frac{du}{u(\ln u)(\ln\ln u)} = \int \frac{dw}{w\ln w}.

Finally, set $z=\ln w$, so $dz=dw/w$. Thus

\int \frac{dw}{w\ln w} = \int \frac{dz}{z} = \ln z+C.

As $x$ tends to infinity, $z=L_3(x)$ tends to infinity, and therefore $\ln z$ is unbounded. The integral diverges, and so does the series. Differentiating $\ln z$ gives $1/z$, confirming the final antiderivative.

Worked Example: A Convergent Third-Level Threshold

Classify

\sum_n \frac{1}{n\,L_1(n)L_2(n)L_3(n)^{3/2}}.

The first two exponents equal one, while the third is $3/2$. The threshold rule predicts convergence. In the integral, set $u=L_1(x)$, then $w=L_2(x)$, then $z=L_3(x)$. The substitutions are $du=dx/x$, $dw=du/u$, and $dz=dw/w$, respectively. Therefore

\int^\infty \frac{dx}{xL_1(x)L_2(x)L_3(x)^{3/2}} = \int^\infty z^{-3/2}\,dz = \left[-2z^{-1/2}\right]_{z_0}^{\infty} = \frac{2}{\sqrt{z_0}},

where $z_0>0$ is the value at the chosen lower limit. This is finite, so the series converges. The derivative of $-2z^{-1/2}$ is $z^{-3/2}$, verifying the antiderivative and the comparison integral.

Worked Example: The First Exponent Already Decides

Classify

\sum_n \frac{1}{n(\ln n)^2\ln\ln n}.

The first exponent is $p_1=2>1$, so the series converges without needing a critical threshold at a later logarithm. This conclusion can also be checked by comparison: for $x\ge e^e$, $\ln\ln x\ge1$, and hence

0< \frac{1}{x(\ln x)^2\ln\ln x} \le \frac{1}{x(\ln x)^2}.

The comparison integral is finite, since

\int_{e^e}^{\infty}\frac{dx}{x(\ln x)^2} = \int_e^\infty \frac{du}{u^2} = \left[-\frac{1}{u}\right]_e^\infty = \frac{1}{e}.

Here $u=\ln x$ and $du=dx/x$. Since the original positive terms are bounded above by the terms of a convergent comparison series, the original series converges. The antiderivative check is $\frac{d}{du}(-1/u)=1/u^2$.

A Common Pitfall: Losing a Logarithm in a Substitution

When substituting $w=\ln u$, the differential $du/u$ becomes $dw$. It is a mistake to treat every remaining logarithmic factor as though it disappears as well. For instance, in the all-critical three-log integral, the substitution $u=\ln x$ gives

\int \frac{dx}{xL_1(x)L_2(x)L_3(x)} = \int \frac{du}{u(\ln u)(\ln\ln u)}.

Incorrectly replacing this immediately by $\int dw/w$ skips a factor. With $w=\ln u$, the factors $\ln u$ and $\ln\ln u$ become $w$ and $\ln w$, respectively, while $du/u=dw$. The correct result is

\int \frac{du}{u(\ln u)(\ln\ln u)} = \int \frac{dw}{w\ln w},

not $\int dw/w$. One more substitution, $z=\ln w$, gives $\int dz/z$, which diverges. Keeping track of both the transformed differential and every remaining denominator factor prevents a wrong convergence conclusion.

How to Apply the Thresholds

Write the exponents in order.
Record the powers of $\ln n,\ln\ln n,\ln\ln\ln n,\ldots$ as $p_1,p_2,p_3,\ldots$.

Check the first exponent.
If $p_1>1$, the series converges; if $p_1<1$, it diverges. If $p_1=1$, continue to $p_2$.

Continue only through critical values.
At each level, an exponent greater than one gives convergence and one less than one gives divergence. Move on only when the exponent equals one.

Check the last boundary carefully.
If every exponent is one, the series diverges. At the final factor, convergence requires an exponent strictly greater than one.

This method matters because logarithmic factors can look similar while producing different outcomes at a critical boundary. A change from exponent $1$ to $3/2$ at the first noncritical level switches divergence to convergence; exponent $1$ itself remains on the divergent side if it is the last factor. Always identify which logarithm is being tested before making the comparison.

Check Your Understanding

For each series, identify the first exponent that differs from one and state whether the series converges or diverges. Assume all sums begin sufficiently far out.

$\sum_n 1/[n\ln n(\ln\ln n)^2]$
$\sum_n 1/[n\ln n(\ln\ln n)^{1/2}]$
$\sum_n 1/[nL_1(n)L_2(n)L_3(n)^{5/4}]$
$\sum_n 1/[nL_1(n)L_2(n)L_3(n)]$
Why does the substitution $w=\ln u$ change $\int du/[u(\ln u)(\ln\ln u)]$ into $\int dw/[w\ln w]$, rather than $\int dw/w$?

Iterated Logarithmic Thresholds

What You'll Learn

Why Iterated Logarithms Create Successive Thresholds

Notation and the Threshold Rule

Why the Successive Substitutions Work

Worked Examples at Different Levels

Worked Example: A Convergent Series at the Second Logarithm

Worked Example: The Final Critical Exponent Diverges

Worked Example: A Convergent Third-Level Threshold

Worked Example: The First Exponent Already Decides

A Common Pitfall: Losing a Logarithm in a Substitution

How to Apply the Thresholds

Check Your Understanding

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