Introduction
An infinite sum of the form
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(1) |
is known as an infinite series. Such series appear in many areas of modern mathematics. Much of this topic was developed during the seventeenth century. Leonhard Euler continued this study and in the process solved many important problems. In this article we will explain Euler’s argument involving one of the most surprising series.
You are likely to have already met perhaps the most important series which is the geometric progression. Given constants and
we want to sum
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(2) |
If we can make sense of the infinite sum – something known by Newton – which is
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(3) |
This was one of the first, and only, general results known during the seventeenth century. Another series then known was
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(4) | ||
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(5) | ||
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(6) | ||
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(7) |
Similar methods were used to find the sums
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(8) |
Now all these series converge. That is to say we can make sense of the infinite sum as a finite number. This is not true of a particularly famous series which is known as the harmonic series:
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(9) |
The following medieval proof that the harmonic series diverges was discovered and published by a French monk called Orseme around 1350 and relies on grouping the terms in the series as follows:
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(10) | ||
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(11) | ||
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(12) |

The harmonic series diverges
It follows that the sum can be made as large as we please by taking enough terms. In fact this series diverges quite slowly. A more accurate estimate of the speed of divergence can be made using the following more modern proof. This uses a technique known as the integral test which compares the graph of a function with the terms of the series. By integrating the function using calculus we can compare the sum of the series with the integral of the function and draw conclusions from this.
In this case we compare terms in the series with the area under the graph of the function . In particular, figure 1 shows that
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(13) |

Figure 1: The series 1/n above the graph of 1/(1+x)
Of course the integral on the right is easy. Solving this gives
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(14) |




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(15) |

Figure 2: The series 1/n below the graph of 1/x
The harmonic series generalised
The harmonic series can be described as "the sum of the reciprocals of the natural numbers". Another series that presents itself as being similar is the "the sum of the squares of reciprocals of the natural numbers". That is to say, the series![]() |
(16) |
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(17) |
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(18) |
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(19) |

Bust of Leibniz by Johann Gottfried Schmidt
"Infinite polynomial" - power series
Before solving this problem we look briefly at a piece of theory Euler used which allowed him to write the function

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(20) |





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(21) |
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(22) |



Euler's solution to the Basel Problem

Leonhard Euler
Euler was working on the Basel Problem at the age of 24 in 1731 by calculating a numerical approximation. This is an arduous task by hand with a series which converges as slowly as this. In 1735 he arrived at the following exact result:
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(23) |
This is a truly remarkable result. No one expected the value , the ratio of the circumference of a circle to the diameter, to appear in the formula for the sum. Euler starts with an
th degree polynomial
with the following properties:
has non-zero roots
,
.
Then may be written as a product in the following form:
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(24) |
We paraphrase Euler’s next claim as "what holds for a finite polynomial holds for an infinite polynomial". He applies this claim to the polynomial
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(25) |
which is an infinite polynomial with . Furthermore, as Euler knew,
can be written as a series:
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(26) |
Multiplying by
he obtained
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(27) |
This has zeros at for
since these are the zeros of
. We can now use the claim above and write
as an infinite product and equate the two as
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(28) | ||
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(29) | ||
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(30) |
The second line pairs the positive and negative roots – the last line uses the difference of two squares to combine these. If you don’t believe this can be done you are right to question the logic here! Euler is being incredibly bold in his assertion that "what holds for a finite polynomial holds for an infinite polynomial". His use turns out to give the correct answer in this case!
Euler’s trick is to write in two different ways. He exploits this by expanding the right hand side. This infinite product will be very complicated but there will be a constant term
and one can collect the
term without too much effort as follows:
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(31) |
Now Euler equates the coefficients of to conclude that
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(32) |
which gives
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(33) |
Now Euler didn’t stop here – he expanded the product further and equated other coefficients to sum other series. In this way he obtained
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(34) |
In 1744 he obtained
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(35) |
by this method. In principle his method solves
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(36) |
for all natural numbers .
Extensions of the Basel problem
In a style typical of Euler, he not only solved the problem in hand but also used the method to solve a class of related problems. You will notice that his method only works for even powers. What then, about![]() |
(37) |
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(38) |
Further reading
You can find out more about some on Euler's work on infinite series (including a derivation of the last result) in his paper Remarques sur un beau rapport entre les series des puissances tant directes que reciproques.
About the author
Chris Sangwin is a member of staff in the School of Mathematics and Statistics at the University of Birmingham. He is a Research Fellow in the Learning and Teaching Support Network centre for Mathematics, Statistics, and Operational Research. His interests lie in mathematical Control Theory.
Comments
Thanks!
Thanks very much for the detailed explanation. It was very helpful.
Hope that I can live until
Hope that I can live until they discover the solution for any zeta function.
Nah, maths develop slowly,
Nah, maths develop slowly, you not going be alive when new maths came out.
Hello
Hi Chris!!!
interesting!
interesting!
Basel
Nice article - i had not seen Euler's method before, thank you.
Very well written and
Very well written and interesting. Regards
Extension to the Basel Problem is known since 1979
Great article. Just a minor hint:
The sum for 1/( k^3 ) is known since 1979:
It's called the Apér'y Constant.
Source:
http://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
Cheers,
Gerhard Tscheinig
Thanks!
Hi Gerhard,
Thanks for pointing us to that info and the wikipedia link. We didn't know the infinite sum for
had a special name.
The sum has been known for a long time, but as far as we know there is still no solution of it in terms of other known numbers, as there is for
Best wishes,
The Plus Team
sorry ,but apery only proved
Sorry, but Apery only proved that it is irrational and that too by following the methods of Leonhard Euler and still the value of this infinite series in unknown. The value has as such been called Apery constant but it is still an open problem for great minds to solve
Partial sum of the basel problem
Hi!
First of all I just wanted to say this is all very well written so thanks.
I am in particular searching for a closed solution for partial sum of the basel problem but can't find it :-(
Ramesh India
Thank you Chris Sangwin
This material is useful.
figure 2 caption
Nice article.
Tiny point.
Figure 2 caption should be ...below the graph of 1/x.
Thanks for spotting that!
Thanks for spotting that! We've corrected it.
Euler, Basel Problem, Newtonian formulae.
Euler solves the Basel problem by applying the Newtonian formulae for converting an infinite summation series into an infinite product series, and vice versa. The Newtonian formulae are explained on pages 358-359 of D.T.Whiteside's Mathematical Papers of Isaac Newton vol 5. This comment submitted by Peter L. Griffiths.
Thank you!
It was really interesting! I think there is also a solution with fourier series.
Zeta evaluations
It is important to mention Euler's discovery in 1755 that the cotangent series generates the zeta evaluations. comment submitted by Peter L. Griffiths.
Great, think there's an error though
On line (?) 26, i think it should be
(-1)^k
instead of
(-1)^k+1
Not sure if i'm right or just missing something...
anyway very interesting and useful article thank you
Thanks for pointing that out!
Thanks for pointing that out! We've fixed the error.
Are equation (24) and
Are equation (24) and equation (25) equal ?
I don't understand the Euler's claim in equation (25).
Hi! No, the two formulae are
Hi! No, the two formulae are not equal as (24) is finite and (25) is infinite. The way Euler's claim is applied is explained in the text between equations (27) and (28): xp(x) is a polynomial with certain roots, whch as in the finite case, can be written as a product.
Eqn 24
Hi, stupid question probably but where does eqn 24 come from!? It doesn't look right. Suppose for example you take p(x) = x^2 - 4, then (24) says p(x) = (1-x/2)(1+x/2), but it doesn't (multiplying out you get 1 - x^2/4, which is not p(x)).
Point (2) says that p(0)=1,
Point (2) says that p(0)=1, which isn't the case for p(x)=x^2-4.
for any equation p(x)=0
for any equation p(x)=0 (where p(x) is an expression or function of x) has roots a0, a1,....then at x=a0, a1,... p(x) becomes 0. in your example, p(x)=1 - x^2/4. At x=2 or -2, i.e., p(2)=p(-2) = 0.
Answer to eq24
Er,it seems your polynomial p(x)=x*2-4 doesn't agree with the condition p(0)=1
Thank you
for the excellent and beautifull explanation
Detail on "assumptions"
Where can one find more detail about Eulers assumptions he made in this proof and how his assumptions were taken care of later?
Euler's Basel Conjecture
The proof of the sine Basel conjecture (PI)^2/2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 .... depends on the Newtonian Infinite Series formulae which are
the ABC summation 1 + Ax + Bx^2 + Cx^3 .... = (1+ax)(1+ bx)(1 +cx)....
the ABC Alternating 1 -Ax + Bx^2 - Cx^3 .... = (1- ax)(1- bx) (1- cx) ...
the A summation Ax = ax + bx + cx..... which seems to be a special case of the sine Basel conjecture (PI)^2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 .....
with both sides multiplied by (u/PI)^2 and thus becomes u^2/6 = (u/PI)^2 + (u/2PI)^2 + (u/3PI)^2 .... clearly a version of the A summation
Au^2 = au^2 + bu^2 + cu^2 ... with A = 1/6, a =(1/PI)^2, b = (1/2PI)^2, c = (1/3PI)^2.
The alternating sine series is (sinu)/u = 1 - u^2/3! + u^4/5! - u^6/7!.....clearly a special case of the ABC Alternating, whose product series can now be
evaluated by substituting values for letters.
expression used for such problems
x=(1/e)^(1/e)^(1/e)...............
What is the value of x
lim[x=infinite]((1/e)^(1/e)^x
lim[x=infinite]((1/e)^(1/e)^x))
powers of power are multiplied.
answer
find by applying ln
Odd positive integer zeta numbers
Hi Chris:
Two years ago I discovered some interesting equations involving odd positive integer zeta values, and recently I derived some equations using Euler's L-number equations (which sum to odd powers of Pi). I believe I am the first person ever to derive these equations. I've been trying to find a closed form for zeta 3 for the last two years since I retired. I've written a few of these equations and derivations using Latex.
Would you be interested in seeing some of my work?
Pete Kacensky
peterkacensky@charter.net
derivation of zeta(4)
can you please show us how to obtain zeta(4) from your presentation of zeta(2)?
Fascinating story told
Fascinating story told fascinatedly