# Euler Series question and solution

Question:

Mengoli had posed the following series to be evaluated:

$1+ \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \ldots$.

Some great mathematicians, including Liebnitz, John Bernoulli and D’Alembert, failed to compute this infinite series. Euler established himself as the best mathematician of Europe (in fact, one of the greatest mathematicians in history) by evaluating this series initially by a not-so-rigorous method. Later on, he gave alternative and more rigorous ways of getting the same result.

Prove that the series converges and gets an upper limit. Then, try to evaluate the series.

Proof:

Due Nicolas Oresine:

Consider the following infinite series: $\phi(s)=1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + ldots$

We can re-write the preceding series as follows: $\phi(s) = 1+ (\frac{1}{2^{s}}+\frac{1}{3^{s}}) + (\frac{1}{4^{s}} + \frac{1}{5^{s}} + \frac{1}{6^{s}} + \frac{1}{7^{s}}) + \ldots$, which in turn is less than

$1 + (\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots$. Now, the RHS of this can be re-written as

$1+(\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots=1 + \frac{1}{2^{(s-1)}}+ (\frac{1}{2^{(s-1)}})^{2} + \ldots$, which is a geometric series and it is given by

$\frac{1}{1-\frac{1}{2^{(s-1)}}}$.

Now, we can say that $\phi(s)$ will converge if $\frac{1}{2^{(s-1)}}<1 \Longrightarrow s >1$.

In order to prove what is asked, we start with $\phi(s)=1 + \frac{1}{2^{s}}+ \frac{1}{3^{s}}+ \frac{1}{4^{s}}+\ldots$

And, then multiply both sides by $\frac{1}{2^{s}}$ and then subtract the resulting equation from the preceding equation to get

$(1-\frac{1}{2^{2}})\phi(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\ldots$

where all the terms containing the reciprocals of the sth power of even numbers vanished.

Repeating this procedure with $\frac{1}{3^{s}}$ gives

$(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})\phi(s)=1+\frac{1}{5^{s}}+ \ldots$

where all terms containing the reciprocals of the sth power of multiples of 3 vanished.

By continuing this with all prime numbers, we get

$\prod_{p}(1-\frac{1}{p^{s}})\phi(s)=1$, where p represents all prime numbers. Thus, we get

$\phi(s)=1 + \frac{}{} + \frac{}{} + \frac{}{} + \ldots =\frac{1}{\prod_{p}(1-\frac{1}{p^{s}})}$

This is a remarkable result because the LHS is concerned with only positive integers, whereas the RHS is concerned with only primes. This result is known as the “Golden Key of Euler”.

Riemann created his famous $\zeta-$ function by extending the variable s to the entire complex plane, except $s=1$ with

$\zeta(s)=1+ \frac{1}{2^{s}} + \frac{1}{3^{s}} + \ldots$.

This function is now very famous as the Riemann zeta function.

How can we apply the Golden Key of Euler to Mengoli’s question that we started with?

Ans. In the Golden Key of Euler, substitute $s=2$.

Hence, we get the upper limit of the given series is 2.

Euler’s proof (1775):

The proof ran as follows:

It is a little roundabout way of arriving at the correct answer from a known result. Consider McLaurin’s series expansion of sin x:

$\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} -\frac{x^{7}}{7!} + \frac{x^{9}}{9!} + \ldots$

By dividing both sides by x and then substituting $y=x^{2}$ on the right side, we get the following:

$\frac{\sin{(x)}}{x} = 1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{7!} + \ldots$

By taking a special value of $x=n\pi$ (and, hence $y=n^{2}\pi^{2}$), we get the following:

$\frac{\sin (n\pi)}{(n\pi)}=0=1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{5!}+ \ldots$

Note  that preceding equation is not a polynomial, but an infinite series. But, Euler still treated it as a polynomial (that is why it was not accepted as a rigorous result) and observed that this “infinite” polynomial has roots equal to $y_{n}=n^{2}x^{2}$. Then, Euler had used the fact that the sum of the reciprocals of the roots is determined by the coefficient of the linear term (here, the y-term) when the constant is made unity. (check this as homework quiz, for a quadratic to be convinced). So, Euler had arrived at the following result:

$1-\sum_{i=1}^{\infty}\frac{6}{y_{n}}=0 \Longrightarrow \sum_{i=1}^{\infty}\frac{1}{y_{n}}=\frac{1}{6}$. With $y_{n}=n^{2}(\pi)^{2}$, we get the following:

$\sum_{i=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}}=\frac{1}{6}$ or, $\sum_{1}^{n^{2}}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}$.

Another proof also attributed to Euler that uses the series expansion of sin (x) goes as follows below:

$\sin {(x)}$ has roots given by 0, $\pm \pi$, $\pm 2\pi$, $\pm 3\pi$, …So does this polynomial that Euler reportedly constructed:

$x(1-\frac{x^{2}}{(\pi)^{2}})(1-\frac{x^{2}}{(2\pi)^{2}})(1-\frac{x^{2}}{(3\pi)^{2}})\ldots=0$

So, Euler considered the preceding equation to be equivalent to:

$\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots=0$

Then, he had equated the coefficient of $x^{3}$ in both to get the result:

$\sum_{n=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}} = \frac{1}{3!} = \frac{1}{6}$.

Thus, $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}$.

Later on, Euler had provided a few more alternate and rigorous proofs of this result.

Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.
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