2. II
Towards infinity

3. Sequence The reciprocals of the terms of the sequence approach zero.
1/n becomes smaller than every fixed e > 0
But the limit is not reached: 1/n > 0

4. Series = sequence of partial sums
Archimedes ( ) knew:
Arithmetic series:
Series of squares:
Geometric series:

5. Carl Friedrich Gauß (1777-1855)
as a school boy: =
n = n(n+1)/2 =

6. (k+1)3 = k3 + 3k2 + 3k + 1
(k+1)3 - k3 = 3k2 + 3k + 1
= 312 + 31 + 1
= 322 + 32 + 1
= 332 + 33 + 1 =
(n+1)3 - n3 = 3n2 + 3n + 1
Blaise Pascal
( )
(n+1) = 3Sk2 + 3Sk + n
3Sk2 = n3 + 3n2 + 3n n(n + 1)/2 - n
Sk2 = n(n + 1)(2n + 1)/6
With Sk2 find Sk3 and so on.

7. k(k+2)(2k+2) = (k-2)k(2k-2) + 12k2
1  3  = 12
3  5  = 1  3  32
5  7  = 3  5  52 =
r(r+2)(2r+2) = (r-2)r(2r-2) + 12r2
Leonardo de Pisa
( )
Fibonacci
r(r+2)(2r+2) = 12( r2)
and analogously
s(s+2)(2s+2) = 12( s2)

8. Geometric series: 1 + q + q2 + q3 + ... + qn
(1 + q + q qn-1 + qn)q
= qn+1
1 + q + q qn =
The inventor of chess
= 21019 grains of rice
Surface of earth: 51018 cm2
1 + q + q → for IqI < 1
Infinitely many numbers, finite sum:

9. The geometric series and its limit for |x| < 1
When writing a bit carelessly
usually the latter is meant
but in fact the former is expressed!

11. Nicole d’Oresme ( )
College de Navarre at Paris:
Pupil, teacher, chairman
Bishop of Lisieux
First ideas of analysis
Rational powers:
43 = 64 = 82  8 = 43/2

12. How long will it take a super computer, to reach the sum S = 100?

13. How long will it take a super computer, to reach the sum S = 100?
k =  S  7
k = 106  S  14
k = 109  S  21
k = 1012  S  28

14. How long will it take a super computer, to reach the sum S = 100?
k =  S  7
k = 106  S  14
k = 109  S  21
k = 1012  S  28
S = 100  k = 1043
With 106 additions per second
1037 seconds will be needed.
The age of the universe  1017 s.
times the age of the universe

15. < 80 When removing all terms containing the digit 9, the
series converges. (Kempner, 1914)
The number of n-digit denominators with no 9 is 89n−1 because there are 8 choices (1 through 8) for the first digit, and 9 choices (0 through 8) for each of the other n−1 digits.
Each of these fractions is at most 10n−1, so the contribution of this group to the sum of fractions is less than 8(9/10)n−1. Therefore the whole sum of reciprocals is at most
Aubrey J. Kempner
( )

16. When removing all terms containing the digit 9, the series converges to about 23. (Frank Irwin ( ) in 1916)
When removing all terms containing the digit string 4711, the series converges.
Proof: Use base 10,000.
Almost all terms contain a 9 like every other string of digits.

17. Not every series converges absolutely:
halved
and added
But we see the same terms!

18. Gottfried Wilhelm Leibniz (1646-1716)
deliberately used the infinite harmoic series:
"... so the difference between two harmonic series, may they be infinite though, can be a finite magnitude."

19. Francois Viète (1540-1603) = Vieta
Born and died as a Catholic,
in between Huguenot.
1572 Massacre of St. Bartholomew:
20000 Huguenots killed.
Attorney in Fontenay-le-Comte.
Member of parliament in Rennes and Tours.
Deciphered the Spanish secret code (500 symbols).
Greatest French mathematician of the 16th century.

20. Theorems about roots of polynomials
x1 + x2 + x xn = -an-1
x1x2x3xn = (-1)na0
follow from the linear factor decomposition
(x - x1)(x - x2)(x - xn) = 0
of the polynomial
xn + an-1xn a1x + a0 = 0
algebra: vocals = unknown, consonant = constant
used +, -
alternative solution of the cubic equation
calculated 9 digits of p
sin2x = 2 sinx cosx and similar formulas

21. sin(x) = 2cos(x/2)sin(x/2)
= 4cos(x/2)cos(x/4)sin(x/4)
= 8cos(x/2)cos(x/4)cos(x/8)sin(x/8)
...
= 2ncos(x/2)cos(x/2n)sin(x/2n)
= x[2n/x]cos(x/2)cos(x/2n)sin(x/2n)
= x cos(x/2)cos(x/2n)sin(x/2n)/[x/2n]
→ 1

22. sin(x) = 2cos(x/2)sin(x/2)
= 4cos(x/2)cos(x/4)sin(x/4)
= 8cos(x/2)cos(x/4)cos(x/8)sin(x/8)
...
= 2ncos(x/2)cos(x/2n)sin(x/2n)
= x[2n/x]cos(x/2)cos(x/2n)sin(x/2n)
= x cos(x/2)cos(x/2n)sin(x/2n)/[x/2n]
sin(x)/x = cos(x/2)cos(x/4)cos(x/8) cos(x/2n) 
sin(p/2)/(p/2) = cos(p/4)cos(p/8)cos(p/16) 
2/p = cos(p/4)cos(p/8)cos(p/16) 
First infinite product-sequence (1593)

23. cos(x) = cos2(x/2) – sin2(x/2)

24. Wallis' Produkt, Arithmetica Infinitorum (1655)
John Wallis
( )

29. James Gregory could calculate all logarithms of positive integers, found the Taylor-series long before Taylor, found 1671 the series of Leibniz, 3 years before Leibniz.
James Gregory
( )

31. Telescope formulas:

32. Illuminated kite
Furniture for doll's house
Prism: dispersion of light
Hairpin-experiment
Reflector telescope improved
Apple tree?
1/r2-law of gravitation
Mechanics: F = p
Isaac Newton
( )

33. Illuminated kite
Furniture for doll's house
Prism: dispersion of light
Hairpin-experiment
Reflector telescope improved
Apple tree?
1/r2-law of gravitation
Mechanics: F = p
Isaac Newton
( )

34. Cambridge Trinity College

35. ¥ Lemniscate Jakob Bernoulli (1654-1705) (1696)
Monk Grandi: So happened God‘s creation from nothing:
0 = = = ½

36. Pupil of Johann Bernoulli.
Greatest mathematician of the 18th century.
Most productive mathematician of all times.
His works fill 70 thick books.
Euler's angles (rigid body)
Euler's gyroscope equations
Euler's buckling equation
Theory of Moon, building of ships, artillery
Labelling of triangles
eij = cosj + isinj eip = -1 (1740)
ln(-1) = ip + ik2p
Leonhard Euler
( )

37. - Son of a clergyman, Paul Euler, who was able to understand the lessons of Jacob Bernoulli.
- Study of theology and mathematics. Favourite student of Johann Bernoulli, the brother and successor of Jacob Bernoulli.
- Wins a prize for a paper on shipbuilding. Very theoretical.
- Sudies one semester medicine because there was a vacation for medicine in St. Petersburg.
- Appointment to St. Petersburg as adjunct for mathematics. But the Tsarina died when Euler entered Russian soil,17. Mai 1727.
- Euler had to serve as a lieutenant in the navy.
Tsarina Anna revives the academy of sciences. Euler works as a mathematician.
Euler loses eyesight of his right eye.
Appointment to the Prussian academy of sciences. Three weeks ocean voyage.
- Euler becomes head of the mathematical and natural scienes class of the Prussian academy.

38. Friederike von Brandenburg-Schwedt
Popuar book: Letters to a German princess.
Translated into 7 languages.
The elevation of Berlin is higher than that of Magdeburg, because the Spree flows into the Havel and this into the Elbe.
But far below of Magdeburg!
Friederike von Brandenburg-Schwedt
( )

39. Tsarina Catherine I
(1684 – )
2nd wife of Peter, the Great
Engaged Euler for the Russian academy of sciences.

40. Friedrich II, the Great
( )
Engaged Euler for the Prussian academy of sciences

41. Sophie von Anhalt-Zerbst
Tsarina Catherine II, the Great
( )
called Euler back.

42. - Euler wins many important awards endowed with lots of money, further he receives a rich pension from Russia.
- Great house and estate, household including 18 persons.
- Johann Bernoulli, well-known as not suffering from too little self-confidence, named him mathematicorum princeps.
- Jean d'Alembert called him diable d'homme.
- But Euler has to do much work unrelated to his position.
Translation of intercepted Russian messages during war.
Constructing fountains in the garden of Sanssouci.
Measuring the Oderbruch in order to drain it.
Calculating pensions.
- Friedrich II is not very satisfied.
Fountains in the garden of Sanssouci don't work.
66-square Greek-Latin-square cannot be presented.
- He held Euler not in high esteem.
- Euler returned to Russia, he was given a reception like a king.

43. Greek-Latin-Squares
All possible combinations of letter and number (or of Latin character and Greek character) have to be realized. All letters and all numbers in every line.
The Prussian army had 6 arms of service and six ranks of officers. They should greet Friedrich II in a Greek-Latin-Square.
But Euler could neither find a 66-square nor prove its non-existence. Euler suspected: (2n+2)-squares don't exist.
Today we know: Except for bases 2 and 6, all exist.

44. Königsberg bridge problem (1741) Is it possible to pass all seven bridges over the Pregel once without using one bridge twice? No. For proof use a rope.
Polyhedron theorem for convex polyhedra (originally stated by Descartes)
Vertices + Sides = Edges + 2
V + S = E + 2
Tetrahedron: = 6 + 2
Cube: =

45. Open the polyhedron and remove a side
Open the polyhedron and remove a side. For the remaining elements we have
V + S' - E = 1.
Cut all sides into triangles. With every generated triangle E and S' grow by 1. V = const.
Open a side touching the hole. E and S decrease by 1.
Remove a free edge. V and E decrease by 1. S = const.
In the end one vertex remains: = 1

46. Leonhard Euler
( )
Infinity (abbreviated by i or the lying S, later by ) is a "size which is realized by increases accumulated without end".
2 (+1)/ log

47. Leonhard Euler
( )
like Leibniz and Jakob Bernoulli.
Assumed with Wallis: 1/3 < 1/2 < 1/1 < 1/0 < 1/-1

48. Euler, like his contemporaries, used divergent sequences for calculations. But he gives the first criterion of convergence: The rest after the infinite term must become infinitely small.
Euler always notes the last term, mostly i for numerus infinitus.

49. n! = 123n generalized for all real numbers.
G-function. Second Euler-integral.

50. The sum of reciprocals 1/p of primes p < N is about lnlnN.
Euler found the longest sequence of primes (at his time):
n(n+1) + 41 supplies primes for n = 0 till n = 39
41, 43, 47, 53, 61, 71, 83,
Analytical proof of the infinity of prime numbers:

51. Every product 1/n appears exactly once.
A finite product would not diverge. (1737)

52. Zeta function z(n)
Starting off from the series for sinx he summed the series of inverse squares. Leibniz and the Bernoullis had tried it in vain.
(Basel problem)
...
Euler failed to find the sums for odd exponents.
Nobody else succeeded yet.
Roger Apéry
( )
Except for the proof by Apéry in 1979 that z(3) = 1, is irrational.

53. for a0 ≠ 0
Zeros of the function sinx are 0, p, 2p, 3p, ...
(1 – x/np)(1 + x/np) = 1 - x2/n2p2

54. Comparison of the quadratic terms supplies:
(1748)
Comparison of higher powers yields higher z-series.

55. and by analytical continuation:
(Applied in string theory!)
...
...

56. Analytical continuation
Let F(z) be an analytical function defined in the nonempty open set U of the complex plane.
Let f(z) be an analytical function defined in the nonempty open subset V of U.
If f(z) = F(z) for all z  V, then F is called the analytic continuation of f.
Example: The Gamma-function G(n) = (n-1)! extended by Euler to all real numbers (except the non-positive integers) has an analytical continuation (a complex integral) to all complex numbers (except the non-positive integers).

57. Example: The Gamma-function G(n) = (n-1)
Example: The Gamma-function G(n) = (n-1)! extended by Euler to all real numbers (except the non-positive integers) has an analytical continuation (a complex integral) to all complex numbers (except the non-positive integers)

58. for Re(z) > 1
It is easier to localize the obvious mistake by observing that the geometric series has no extension to x = 1.
Otherwise mathematics would be inconsistent by 1 < Sn < 0.
for |x| < 1

59. for all x with |x| < 1
for x = 1

60. for all x with |x| < 1
for x = 1

61. =
digits are known.
ln(a/b) = lna - lnb

63. Appendix