more proofs of the diverge of the harmonic series

8/10/2019 More Proofs of the Diverge of the Harmonic Series

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More Proofs of Divergence of the Harmonic Series

Steven J. Kifowit

Prairie State College

In an earlier article, Kifowit and Stamps [13] summarized a number of elementary proofs of divergence of the harmonic series:

n =1

1n

= 1 + 12

+ 13

+ 14

+ 15

+ = .

For a variety of reasons, some very nice proofs never made it into the nal draft of that article.With this in mind, the collection of divergence proofs continues here. This informal note is awork in progress 1. On occasion more proofs will be added. Accessibility to rst-year calculusstudents is a common thread that will continue (usually) to connect the proofs.

Proof 21 (A geometric series proof)

Choose a positive integer k.

n =1

1n

= 1 +

k terms

12 + 13 + + 1k + 1 +k 2 terms

1k + 2 + 1k + 3 + + 1k2 + k + 1+

k 3 terms

1k2 + k + 2 + 1k2 + k + 3 + + 1k3 + k2 + k + 1 + > 1 +

kk + 1

+ k2

k2 + k + 1 +

k3

k3 + k2 + k + 1 +

> 1 + k

k + 1+

kk + 1

2

+ k

k + 1

3

+ =

11 kk+1

= k + 1

Since this is true for any positive integer k, the harmonic series must diverge.

1 First posted January 2006. Last updated February 16, 2013.


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Proof 22

The following proof was given by Fearnehough [9] and later by Havil [10]. After substituting u = ex , thisproof is equivalent to Proof 10 of [13].

0 ex1

ex dx = 0 ex (1 ex ) 1 dx

= 0

ex (1 + ex + e2x + e3x + ) dx

= 0 (ex + e2x + e3x + ) dx= ex +

12

e2x + 13

e3x + 0

= 1 + 12

+ 13

+ = [ln(1 ex )]

0 =

Proof 23 (A telescoping series proof)

This proof was given by Bradley [2]. We begin with the inequality x ln(1+ x), which holds for all x > 1.From this it follows that

1k ln 1 +

1k

= ln( k + 1) ln(k)for any positive integer k. Now we have

H n =n

k=1

1k

n

k=1

ln 1 + 1

k =

n

k=1

lnk + 1

k

= [ln( n + 1) ln(n)] + [ln( n) ln(n 1)] + + [(ln(2) ln(1)]= ln( n + 1) .

Therefore {H n } is unbounded, and the harmonic series diverges.

Proof 24 (A limit comparison proof)

In the last proof the harmonic series was directly compared to the divergent series

k=1

ln 1 + 1k

. The use

of the inequality x ln(1 + x) can be avoided by using limit comparison. Since

limx

ln 1 + 1x1x

= limx

1x 21 + 1x 1x 2

= 1 ,

the harmonic series diverges by limit comparison.

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Proof 25

In an interesting proof of the Egyptian fraction theorem, Owings [18] showed that no number appears morethan once in any single row of the following tree.

1/2

1/31/4

1/5...

1/20...

1/12

1/13...

1/156...

1/61/7

1/8...

1/56...

1/42

1/43...

1/1806...

The elements of each row have a sum of 1 / 2, and there are innitely rows with no elements in common. (Forexample, one could, starting with row 1, nd the maximum denominator in the row, and then jump to thatrow.) It follows that the harmonic series diverges.

Proof 26

This proof is actually a pair of very similar proofs. They are closely related to a number of other proofs, butmost notably to Proof 4 of [13]. In these proofs H n denotes the n th partial sum of the harmonic series:

H n = 1 + 12

+ 13

+ + 1n

, n = 1 , 2, 3, . . . .

Proof (A): First notice that

H n + H 2n = 2H n + 1n + 1

+ 1n + 2

+ + 12n 2H n +

n2n

,

so that when all is said and done, we have

H n + H 2n 2H n + 12 .

Now suppose the harmonic series converges with sum S .

2S = limn

H n + limn H 2n

= limn

(H n + H 2n )

limn 2H n + 12

= 2 S + 12

The contradiction 2 S 2S + 12 concludes the proof.

Proof (B): This proof was given by Ward [20].

H 2n H n = 1n + 1

+ 1n + 2

+ + 12n

n2n

= 12

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Suppose the harmonic series converges.

0 = limn

H 2n limn H n= lim

n (H 2n H n )

limn 12

= 12

The contradiction 0 12 concludes the proof.

Proof 27

This proposition follows immediately from the harmonic mean/arithmetic mean inequality, but an alternateproof is given here.

Proposition: For any natural number k, 1

k +

1k + 1

+ + 13k

> 1.Proof:

exp 1k + 1k + 1 + 1k + 2 + + 13k = e1/k e1/ (k+1) e1/ (k+2) e1/ (3 k)

> 1 + 1k 1 +

1k + 1 1 +

1k + 2 1 +

13k

=k + 1

k k + 2k + 1

k + 3k + 2

3k + 13k

= 3k + 1

k > 3.

(In this proposition the denominator 3 k could be replaced by e k, but even this choice isnot optimal. See [1] and [4].)

Based on this proposition, we have the following result:

n =1

1n

= 1 +12

+ + 16

+17

+ + 121

+ 122

+ + 166

+ > 1 + 1 + 1 + 1 +

Proof 28 (A Fibonacci number proof)

The following proof was given by Kifowit and Stamps [12] and by Chen and Kennedy [3].The Fibonacci numbers are dened recursively as follows:

f 0 = 1 , f 1 = 1; f n +1 = f n + f n 1 , n = 1 , 2, 3 . . . .

For example, the rst ten are given by 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. The sequence of Fibonacci numbersmakes an appearance in a number of modern calculus textbooks (for instance, see [14] or [19]). Often thelimit

limn

f n +1f n

= = 1 + 5

2

4


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Proof 30

The following visual proofs show that by carefully rearranging terms, the harmonic series can be madegreater than itself.

Proof (A): This proof without words was posted on The Everything Seminar (Harmonic Digression,http://cornellmath.wordpress.com/2007/07/12/harmonic-digression/ ).

: : .. .

Figure 1: Proof 30(A)

Proof (B): This visual proof leaves less to the imagination than Proof (A). It is due to Jim Belk andwas posted on The Everything Seminar as a follow-up to the previous proof. Belks proof is a visualizationof Johann Bernoullis proof (see Proof 13 of [13]).

: : ...

Figure 2: Proof 30(B)

Proof (C): This proof is a visual representation of Proofs 6 and 7 of [13]. With some minor modications,

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a similar visual proof could be used to show that one-half of the harmonic series ( 12 + 14 +

16 + ) is strictly

less than its remaining half (in the spirit of Proof 8 of [13]).

1

1/21/3 1/4 1/5 1/6 1/7 1/8

11/2 1/3 1/4

Figure 3: Proof 30(C)

Proof 31

Here is another proof in which 1/k and 1/xdx are compared. Unlike its related proofs (e.g. Proof 9 of [13]), this one focuses on arc length.The graph shown here is that of the polar function r () = / on [, ).

x

y

r () =

The total arc length is unbounded:

Arc Length =

r 2 + drd 2 d

r 2 d =

r d =

d = ln

=

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Now let the polar function be dened by

() =

1, < 21/ 2, 2 < 31/ 3, 3 < 4... ...1/n, n < (n + 1) ... ...

The graph of is made up of semi-circular arcs, the n th arc having radius 1 /n . The total arc length of the graph of is

+ 2

+ 3

+ 4

+ = 1 + 12

+ 13

+ 14

+ .The graphs of r (dashed) and (solid) on [, 6] are shown below.

x

y

By comparing the graphs of the two functions over intervals of the form [ n, (n + 1) ], we see that thegraph of must be longer than the graph of r . It follows that

1 + 12

+ 13

+ 14

+ must be unbounded.

Proof 32

The divergence of the harmonic series follows immediately from the Cauchy Condensation Test:

Suppose {an } is a non-increasing sequence with positive terms. Then

n =1an converges if and

only if

k=0

2k a2k converges.

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Proof 33

This proof is similar to Proof 4 of [13]. Just as above, H n denotes the nth partial sum of the harmonic series:

H n = 1 + 12

+ 13

+ + 1n

, n = 1 , 2, 3, . . . .

Consider the gure shown here:

1n

1n +1

n n + 1

y = 1x

Referring to the gure, we see that

1n + 1

< n +1n 1x dx < 1n .Repeated use of this result gives

1n + 1

+ + 12n

H 2 n H n

< 2nn 1x dx = ln 2 < 1n + + 12n 1

H 2 n 1 H n 1

orH 2n H n < ln 2 < H 2n H n +

1n

12n

1/ 2n.

From this it follows thatln 2

12n

< H 2n H n < ln 2.Therefore H 2n H n ln 2, and the sequence {H n } must diverge.

Proof 34

In [8] Paul Erd os gave two remarkably clever proofs of the divergence of the series 1/p ( p prime), wherethe sum is taken over only the primes. Other proofs of this fact, such as one given by Euler (see [7]), makeuse of the harmonic series. Erd os proofs do not, and as a consequence, they establish the divergence of theharmonic series.

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The proofs are rather complicated, but accessible and well worth the effort required to follow themthrough. A few elementary ideas are required before we begin:

(i) x denotes the greatest integer less than or equal to x.(ii) We denote the primes, in ascending order, by p1 , p2 , p3 , p4, . . . .

(iii) Let N be a given positive integer. For any positive integer m, there are N/m integers between 1 andN that are divisible by m.

This is the rst of Erdos proofs. We begin by assuming

i =1

1 pi

converges. It follows that there exists an

integer K such that

i = K +1

1 pi

< 12

.

We will call pK +1 , pK +2 , pK +3 , . . . the large primes, and p1 , p2 , . . . , pK the small primes.Now let N be an integer such that N > p K . Let N 1 be the number of integers between 1 and N whose

divisors are all small primes, and let N 2 be the number of integers between 1 and N that have at least one

large prime divisor. It follows that N = N 1 + N 2 .By the denition of N 2 and using (iii) above, it follows that

N 2 N

pK +1+

N pK +2

+ N

pK +3+ =

i= K +1

N pi

.

From this we get

N 2

i = K +1

N pi

i = K +1

N pi

< N

2 ,

where the last part of the inequality follows from the denition of K .Now, referring back to the small primes, let x

N be a positive integer with only small prime divisors.

Write x = yz2 , where y and z are positive integers, y is squarefree (i.e. has no perfect square divisors), andz N . The integer y must have the factorization

y = pm 11 pm 22 p

m 33 pm KK ,

where each exponent mi has value 0 or 1. It follows from the multiplication principle that there are 2 K

possible choices for the integer y. Since z N , there are at most N possible choices for the integer z.Therefore there are at most 2 K N possible choices for the integer x. Recalling the denition of x, we seethat we must have N 1 2K N .

So now we have established that

N = N 1 + N 2 < 2K N + N 2 .However, by simply choosing N such that N > 22K +2 , we are lead to a contradiction:

N = N 1 + N 2 < 2K N + N 2 < N

2 N + N

2 = N.

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Proof 35

This is Erd os second proof of the divergence of 1/p ( p prime) [8]. It uses the same notation and conceptsas the previous proof.

We begin by using the fact that

i=2

1i(i + 1) =

i =2

1i

1i + 1 =

12

to establish that

i =1

1 p2i

< 14

+ 12 3

+ 13 4

+ 14 5

+ = 14

+ 12

= 34

.

Now assume that

i=1

1 pi

converges. It follows that there exists an integer K such that

i = K +1

1 pi

< 18

.

As above, we will call pK +1 , pK +2 , pK +3 , . . . the large primes, and p1 , p2 , . . . , pK the small primes.Let N be a positive integer and let y N be a positive, squarefree integer with only small prime divisors.

The integer y must have the factorization

y = pm 11 pm 22 p

m 33 pm KK ,

where each exponent m i has value 0 or 1. It follows from the multiplication principle that there are 2 K possi-ble choices for the integer y. Those 2k integers must remain after we remove from the sequence 1 , 2, 3, . . . , N all those integers that are not squarefree or have large prime divisors. Therefore, we must have the followinginequality:

2K N K

i =1

N p2i

i= K +1

N pi N

K

i=1

N p2i

i= K +1

N pi

> N 34N 18N = N 8 .

However, if we simply choose N 2K +3 , we have a contradiction.

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Proof 36

Nick Lord [15] provided this visual catalyst for Proof 23.Consider the graph of y = sin( ex ) for 0 x < .

x

y y = sin( ex )

The graph has an x-intercept at the point where x = ln n for each positive integer n . This sequence of x-values diverges to innity, and the distance between each pair of intercepts is given by

ln(n + 1) ln n = lnn + 1

n = ln 1 +

1n

.

Since ln(1+ 1n ) < 1n for each n, the series of gaps between intercepts has a total length less than the harmonic

series:

n =1ln 1 +

1n

10 1 xx dx = Proof 38

Here is a matrix version of Johann Bernoullis proof (Proof 13 of [13]). We start by dening the innitematrices M (square) and h:

M =

0 2 0 0 0 0 0 0 3 0 0 0 0 0 0 4 0 0 0 0 0 0 5 0 ... ... ... ... ... ... . . .

, h =

11/ 21/ 31/ 4

...

Notice that hT Mh =

n =11n = 1 +

12 +

13 +

14 + . Therefore our goal is to show that hT Mh = .

Let J k be the innite square matrix with ones along the superdiagonal starting at row k and with zeros

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elsewhere. For example,

J 3 =

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1

... ... ... ... ... ... . . .

.

It follows that M = J 1 + J 1 + J 2 + J 3 + J 4 + (in the sense that, for any n, the equality holds for the n nleading, principal submatrices).

By expanding hT J 1h, we nd that

hT J 1h =

n =1

1n(n + 1)

= 1 .

From here it is easy to show (perhaps by induction) that

hT J k h = 1

k.

Now suppose that the harmonic series converges with sum S . Then we have

S = hT Mh = hT (J 1 + J 1 + J 2 + J 3 + J 4 + )h= hT J 1h + hT J 1h + hT J 2h + hT J 3 h + hT J 4h + = 1 + 1 + 12 +

13 +

14 +

= 1 + S.

The conclusion, S = 1 + S , is impossible unless S = .

Proof 39 (Cesaro summability)It is not difficult to prove the following result (see for example [11, pages 128129]), which was studied indetail by the Italian mathematician Ernesto Ces` aro.

Suppose that k=1 ak converges with sum S and let S n = nk=1 ak . Then the sequence of average partial sums, {1n nk=1 S k}n =1 , also converges to S .In order to use this result, let

H n = 1 + 12

+ 13

+ 14

+ + 1n

,

and notice that

1n

n

k=1H k = 1n n1 + n 12 + n 23 + + 1n

= 1

n

n

k=1

n k + 1k

= H n 1 + H n

n

.

(A Proof Without Words for this last fact is given in [17].)

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The divergence of the harmonic series now follows by contradiction. Assuming that the harmonic seriesconverges with sum S , we have

S = limn

1n

n

k=1

H k = limn

H n 1 + H n

n = S 1 + 0 .

Proof 40The following proof was given by Augustus De Morgan [6]. Similar to several of the previous proofs, it isincluded here for its historical signicance. De Morgan claims to have been shown this proof many yearsprior to its publication by his young student J. J. Sylvester, who himself went on to become a famousmathematician. De Morgans presentation is duplicated verbatim:

It is well known that when a b + c d + . . . consists of terms diminishing without limit, the series isconvergent, with a limit between a and a b. Now

1 + 12 + 1

3 + 1

4 + . . . is 1 12 + 13 14 + . . .+ 1 + 12 + . . .

And if it be S , we have S = a + S , where a is nite. Hence S is innite.

Proof 41

This proof was given by Cusumano [5]. Suppose m is an integer greater than 1.

n =1

1n

=11

+ + 1m

+ 1

m + 1 + +

1m2

+ 1

m2 + 1 + +

1m3

+ 1

m3 + 1 + +

1m4

+

> m

m +

m 2 mm2

+ m 3 m2

m3 +

m 4 m3m4

+

= 1 + m(m 1)

mm +

m 2(m 1)m2m

+ m 3(m 1)

m3m +

= 1 + m 1

m +

m 1m

+ m 1

m +

= Notice that if the rst term on the right is rewritten,

11

+ + 1m

= 1 +12

+ + 1m

,

Cusumanos proof becomes precisely the generalization of Oresmes classical proof that was described afterProof 1 of [13].

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Proof 42 (Eulers constant)

There are a variety of proofs of divergence of the harmonic series that in some way make use of the Euler-Mascheroni constant. This constant is often dened by the following limit:

= limn

n

k=1

1k ln(n) 0.5772156649.

Once this limit has been established (see for example [14, page 623, exercise 75]), it is clear that the partialsums, H n = nk=1 1k , are unbounded.Here is a different proof involving . Let n be a xed positive integer. For k = 1 , 2, 3, . . ., let

S k = 1k

n

j =1

1kn + j

.

For example,S 1 = 1

1n + 1

1n + 2

12n

,

S 2 = 1

2 1

2n + 1 1

2n + 2 1

3n,

andS n 1 =

1n 1

1(n 1)n + 1

1(n 1)n + 2

1n2

.

Sincen

kn + n

more proofs of the diverge of the harmonic series

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