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On a convergence/divergence problem from integral
calculus
Aziz Contractor∗ and Elliot Krop
Clayton State University
Department of Mathematics
November 1, 2014
Aziz Contractor (Clayton State University) On a convergence/divergence problem from integral calculusNovember 1, 2014 1 / 18

Table of contents
1 A Problem from Calculus 2
2 A Harder Problem
3 A Simple Elegant Surprise
4 References

Infinite Series
One of the first facts about infinite series we learn in second semester
calculus is about the divergence of the harmonic series
∞
n=1
1
n

Inﬁnite Series
∞
n=1
1
n
This is often contrasted with the convergence of
∞
n=1
1
n2

Inﬁnite Series
∞
n=1
1
n
This is often contrasted with the convergence of
∞
n=1
1
n2
Both of these results can be easy applications of the integral test or
the Cauchy Condensation test

Inﬁnite Series
Just in case our memory fails us, here is the statement of that test:

Inﬁnite Series
Theorem
Suppose that an > 0 for any n ≥ 1 and that an is monotone decreasing,
then the series an and 2na2n either both converge or both diverge.

Inﬁnite Series
Theorem
Suppose that an > 0 for any n ≥ 1 and that an is monotone decreasing,
then the series an and 2na2n either both converge or both diverge.
We can further investigate this convergence/divergence phenomenon
by introducing logarithms in the denominator.

Inﬁnite Series
A natural question is to ask how ﬁnely can we increase the
denominator and maintain the same convergence/divergence
behavior. As before, we apply Cauchy’s test we get similar results:
∞
n=1
1
n ln n
diverges

Inﬁnite Series
A natural question is to ask how ﬁnely can we increase the
denominator and maintain the same convergence/divergence
behavior. As before, we apply Cauchy’s test we get similar results:
∞
n=1
1
n ln n
diverges
∞
n=1
1
n(ln n)2
converges

Inﬁnite Series
This can be continued...

Inﬁnite Series
∞
n=1
1
n ln n(ln(ln n))
diverges

Inﬁnite Series
∞
n=1
1
n ln n(ln(ln n))
diverges
∞
n=1
1
n ln n(ln(ln n))2
converges

The Question
It is clear that with any ﬁnite number of logarithms, if we follow the
previous pattern, we have the previous split behavior.

The Question
But what if the number of logarithms continues to increase?

The Question
More precisely, we deﬁne the function
g(k) =



0 : 0 ≤ k < e
1 : e ≤ k < ee
2 : ee ≤ k < eee
...
...
which counts the order of the largest tower function of e, less than or
equal to k.

The Question
More precisely, we deﬁne the function
g(k) =



0 : 0 ≤ k < e
1 : e ≤ k < ee
2 : ee ≤ k < eee
...
...
which counts the order of the largest tower function of e, less than or
equal to k.
Furthermore, say ln(m)
(k) = ln(ln . . . (ln k) . . . ) where the
composition contains m logarithms.

The Question
With this notation, we can let
ak =
1
k
g(k)
i=0 (ln(i)
k)

The Question
With this notation, we can let
ak =
1
k
g(k)
i=0 (ln(i)
k)
and ask:
Does
∞
k=1
ak diverge?

Some Failed Attempts
1 Comparison Tests (Couldn’t ﬁnd the right function for the
comparison)
2 Integral Test (Diﬃcult integral)
3 Cauchy Condensation Test (Very messy)

A Test From the Past
In 1871, Vasilii Petrovich Ermakov, working from the University of
Kiev, showed the following result:
Theorem (Ermakov)
Let f (x) be a positive decreasing function for x ≥ 1.
1 If there exists λ < 1 such that ex f (ex )
f (x) < λ for suﬃciently large x,
then the series f (n) converges.
2 If ex f (ex )
f (x) ≥ 1 for all suﬃciently large x, then the series f (n)
diverges.

Ermakov’s Idea
proof of (1).
First, suppose that lim supx→∞
ex f (ex )
f (x) = L < 1 and choose L1 to be any
number such that L < L1 < 1.

Ermakov’s Idea
proof of (1).
ex f (ex )
By deﬁnition of limit supremum, we can ﬁnd N so that ex f (ex ) < L1 f (x)
for x > N, and by integrating,
x
N etf (et)dt < L1
x
N f (t)dt.

Ermakov’s Idea
proof of (1).
ex f (ex )
x
N etf (et)dt < L1
x
N f (t)dt.
By an easy substitution, this can be rewritten as
ex
eN f (t)dt < L1
x
N f (t)dt
which is the same as
(1 − L1)
ex
eN f (t)dt < L1
x
N f (t)dt −
ex
eN f (t)dt = (∗).

Ermakov’s Idea
proof of (1).
ex f (ex )
x
N etf (et)dt < L1
x
N f (t)dt.
ex
eN f (t)dt < L1
x
N f (t)dt
(1 − L1)
ex
eN f (t)dt < L1
x
N f (t)dt −
ex
eN f (t)dt = (∗).
Adding and subtracting the integral of f between x and eN gives
(∗) < L1
eN
N f (t)dt −
ex
x f (t)dt .

Ermakov’s Idea
proof of (1).
ex f (ex )
x
N etf (et)dt < L1
x
N f (t)dt.
ex
eN f (t)dt < L1
x
N f (t)dt
(1 − L1)
ex
eN f (t)dt < L1
x
N f (t)dt −
ex
eN f (t)dt = (∗).
Adding and subtracting the integral of f between x and eN gives
(∗) < L1
eN
N f (t)dt −
ex
x f (t)dt .
Trivially, since ex > x, (1 − L1)
ex
eN f (t)dt < L1
eN
N f (t)dt for any x > N.
Thus, the integral
∞
a f converges, and so does the series.

Ermakov’s Idea
proof of (2).
For the second statement, we see that one can ﬁnd N such that
ex f (ex ) ≥ f (x) whenever x ≥ N.

Ermakov’s Idea
proof of (2).
As before, we can integrate and substitute to write
ex
eN f (t)dt ≥
x
N f (t)dt
when x > N.

Ermakov’s Idea
proof of (2).
ex
eN f (t)dt ≥
x
N f (t)dt
when x > N.
Subtracting
x
eN f (t)dt from both sides, we obtain,
ex
x f (t)dt ≥
eN
N f (t)dt when x > N.

Ermakov’s Idea
proof of (2).
ex
eN f (t)dt ≥
x
N f (t)dt
when x > N.
Subtracting
x
eN f (t)dt from both sides, we obtain,
ex
x f (t)dt ≥
eN
N f (t)dt when x > N.
This shows that the integral
∞
a f diverges, because for every x,
ex
x f (t)dt
is greater than some fixed constant. Thus, the integral is at least as large
as an infinite sum of that fixed constant, and the sum diverges as well.

Applying Ermakov’s Test
By Ermakov’s test, we can solve our problem by observing
lim
x→∞
ex f (ex )
f (x)
= lim
x→∞
ln(g(x))
(x) (2.1)

lim
x→∞
ex f (ex )
f (x)
= lim
x→∞
ln(g(x))
(x) (2.1)
It is not hard to show that (2.1)≥ 1.

lim
x→∞
ex f (ex )
f (x)
= lim
x→∞
ln(g(x))
(x) (2.1)
Notice, for example, when ee ≤ x < eee
,
ln(ln(x)) = ln(ln(ee
+ )) ≥ 1

lim
x→∞
ex f (ex )
f (x)
= lim
x→∞
ln(g(x))
(x) (2.1)
Notice, for example, when ee ≤ x < eee
,
ln(ln(x)) = ln(ln(ee
+ )) ≥ 1
The same observation holds for x chosen in any interval of tower
functions of e.

A Simple Elegant Surprise
It turns out another solution to this problem was given by R.P. Agnew
of Cornell University in 1947, which he himself attributed to B.
Pettineo published in the national academy of Italy, Accademia dei
Lincei, in 1946.

It turns out another solution to this problem was given by R.P. Agnew
of Cornell University in 1947, which he himself attributed to B.
Pettineo published in the national academy of Italy, Accademia dei
Lincei, in 1946.
The solution is given by a simple sequence of integrals.

Let P(x) denote the product of x and all of the numbers
ln x, ln ln x, ln ln ln x, . . . which are greater than 1.

Consider the following sequence of integrals.
e
1
1
P(x)
dx = ln x
e
1
= 1
ee
e
1
P(x)
dx = ln ln x
ee
e
= 1
eee
ee
1
P(x)
dx = ln ln ln x
eee
ee
= 1

Consider the following sequence of integrals.
e
1
1
P(x)
dx = ln x
e
1
= 1
ee
e
1
P(x)
dx = ln ln x
ee
e
= 1
eee
ee
1
P(x)
dx = ln ln ln x
eee
ee
= 1
Continuing this way produces an inﬁnite set of intervals, over each of
which the integral 1
P(x)dx is 1.

Further Study
Real Inﬁnite Series by D.D. Bonar and M.J. Khoury

Thank You

References
R. P. Agnew, A Slowly Divergent Series, American Mathematical
Monthly, Vol. 54, No. 5, p. 273-274 (1947).
D. D. Bonar and M. J. Khoury, Real Infinite Series, Mathematical
Association of America. ISBN 978-0-8838-5745-8 (2006)
T. J. Bromwich, An Introduction to the Theory of Infinite Series,
American Mathematical Society Chelsea Publishing. ISBN
978-0-8284-0335-1 (1926)
V. P. Ermakov, A new criterion for convergence and divergence of
infinite series of constant sign, Bulletin des Sciences Mathmatiques, t.
2, p. 250, Kiev (1871) (In Russian)
B. Pettineo, Estensione di una classe di serie divergenti. Atti della
Reale Accademia Nazionale dei Lincei, Rendiconti, Classe di Scienze
Fisiche, Matematiche e Naturali. Series 8, vol 1, p. 680-685 (1946).

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