# 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.

Hope you all enjoyed it — learning to think like Euler !! By the way, it did take a long time for even analysis to become so rigorous as it is now….You might like this observation a lot. 🙂 🙂 🙂

Nalin Pithwa.

# An algebra question for RMO or Pre-RMO

Question:

If a, b, c are non-negative real numbers such that $(1+a)(1+b)(1+c)=8$, then prove that the product abc cannot exceed 1.

Solution will be posted after you attempt it…so that you can compare the two approaches.

Nalin Pithwa.

# RMO Training: taking help from Nordic mathematical contest: 1988

Problem:

Let $m_{n}$ be a smallest value of the function $f_{n}(x)=\sum_{k=0}^{2n}x^{k}$. Prove that $m_{n} \rightarrow \frac{1}{2}$ when $n \rightarrow \infty$.

Proof:

For $n>1$,

$f_{n}(x)=1+x+x^{2}+\ldots=1+x(1+x+x^{2}+x^{4}+\ldots)+x^{2}(1+x^{2}+x^{4}+\ldots)=1+x(1+x)\sum_{k=0}^{n-1}x^{2k}$.

From this, we see that $f_{n}(x)\geq 1$ for $x \leq -1$ and $x\geq 0$. Consequently, $f_{n}$ attains its maximum value in the interval $(-1,0)$. On this interval

$f_{n}(x)=\frac{1-x^{2n+1}}{1-x}>\frac{1}{1-x}>\frac{1}{2}$

So, $m_{n} \geq \frac{1}{2}$. But,

$m_{n} \leq f_{n}(-1+\frac{1}{\sqrt{n}})=\frac{1}{2-\frac{1}{\sqrt{n}}}+\frac{(1-\frac{1}{\sqrt{n}})^{2n+1}}{2-\frac{1}{\sqrt{n}}}$

As $n \rightarrow \infty$, the first term on the right hand side tends to the limit $\frac{1}{2}$. In the second term, the factor

$(1-\frac{1}{\sqrt{n}})^{2n}=((1-\frac{1}{\sqrt{n}})^{\sqrt{n}})^{2\sqrt{n}}$

of the numerator tends to zero because

$\lim_{k \rightarrow \infty}(1-\frac{1}{k})^{k}=e^{-1}<1$.

So, $\lim_{n \rightarrow \infty}m_{n}=\frac{1}{2}$

auf wiedersehen,

Nalin Pithwa.

Reference: Nordic Mathematical Contest, 1987-2009.

# An easy inequality from Nordic mathematical contests !?

Reference: Nordic Mathematical Contest, 1987-2009, R. Todev.

Question:

Let a, b, and c be real numbers different from 0  and $a \geq b \geq c$. Prove that inequality

$\frac{a^{3}-c^{3}}{3} \geq abc(\frac{a-b}{c} + \frac{b-c}{a})$

holds. When does the equality hold?

Proof:

We know that a, b and c are real, distinct and also non-zero and also that $a \geq b \geq c$.

Hence, $c-b \leq 0 \leq a-b$, we have $(a-b)^{3}\geq (c-b)^{3}$, or

$a^{3}-3a^{a}b+3ab^{2}-b^{3} \geq c^{3}-3bc^{2}+3b^{2}c-b^{3}$

On simplifying this, we immediately have

$\frac{1}{3}{(a^{3}-c^{3})} \geq a^{2}b-ab^{2}+b^{2}c-bc^{2}=abc(\frac{a-b}{c}+\frac{b-c}{a})$.

A sufficient condition for equality is $a=c$. If $a>c$, then $(a-b)^{3}>(c-b)^{3}$. which makes the proved inequality a strict one. So, $a=c$ is a necessary condition for equality too.

-Nalin Pithwa.

# Algebra : max and min: RMO/INMO problem solving practice

Question 1:

If $x^{2}+y^{2}=c^{2}$, find the least value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}$.

Solution 1:

Let $z^{'}=\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{y^{2}+x^{2}}{x^{2}y^{2}} = \frac{c^{2}}{x^{2}y^{2}}$

$z^{'}$ will be minimum when $\frac{x^{2}y^{2}}{c^{2}}$ will be minimum.

Now, $let z=\frac{x^{2}y^{2}}{c^{2}}=\frac{1}{c^{2}}(x^{2})(y^{2})$….call this equation I.

Hence, z will be maximum when $x^{2}y^{2}$ is maximum but $(x^{2})(y^{2})$ is the product of two factors whose sum is $x^{2}+y^{2}=c^{2}$.

Hence, $x^{2}y^{2}$ will be maximum when both these factors are equal, that is, when

$\frac{x^{2}}{1}=\frac{y^{2}}{1}=\frac{x^{2}+y^{2}}{1}=\frac{c^{2}}{1}$. From equation I, maximum value of $z=\frac{c^{2}}{4}$. Hence, the least value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}=\frac{4}{c^{2}}$.

Some basics related to maximum and minimum:

Basic 1:

Let a and b be two positive quantities, S their sum and P their product; then, from the identity:

$4ab=(a+b)^{2}-(a-b)^{2}$, we have

$4P=S^{2}-(a-b)^{2}$ and $S^{2}=4P+(a-b)^{2}$.

Hence, if S is given, P is greatest when $a=b$; and if P is given, S is least when $a=b$. That is, if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.

Basic 2:

To find the greatest value of a product the sum of whose factors is constant.

Solution 2:

Let there be n factors $a,b,c,\ldots, k$, and suppose that their sum is constant and equal to s.

Consider the product $abc\ldots k$, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors $\frac{a+b}{2}, \frac{a+b}{2}$, the product is increased, while the sum remains unaltered; hence, so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is $\frac{s}{m}$, and the greatest value of the product is $(\frac{s}{n})^{n}$, or $(\frac{a+b+c+\ldots+k}{n})^{n}$.

Corollary to Basic 2:

If $a, b, c, \ldots k$ are unequal, $(\frac{a+b+c+\ldots+k}{n})^{2}>abc\ldots k$;

that is, $\frac{a+b+c+\ldots +k}{n} > (\frac{a+b+c+\ldots + k}{n})^{\frac{1}{n}}$.

By an extension of the meaning of the terms arithmetic mean and geometric mean, this result is usually stated as follows: the arithmetic mean of any number of positive quantities is greater than the geometric mean.

Basic 3:

To find the greatest value of $a^{m}b^{n}c^{p}\ldots$ when $a+b+c+\ldots$ is constant; m,n, p, ….being positive integers.

Solution to Basic 3:

Since m,n,p, …are constants, the expression $a^{m}b^{n}c^{p}\ldots$ will be greatest when $(\frac{a}{m})^{m}(\frac{b}{n})^{n}(\frac{c}{p})^{p}\ldots$ is greatest. But, this last expression is the product of $m+n+p+\ldots$ factors whose sum is $m(\frac{a}{m})+n(\frac{b}{n})+p(\frac{c}{p})+\ldots$, or $a+b+c+\ldots$, and therefor constant. Hence, $a^{m}b^{n}c^{p}\ldots$ will be greatest when the factors $\frac{a}{m}, \frac{b}{n}, \frac{c}{p}, ldots$ are all equal, that is, when

$\frac{a}{m} = \frac{b}{n} = \frac{c}{p} = \ldots = \frac{a+b+c+\ldots}{m+n+p+\ldots}$

Thus, the greatest value is $m^{m}n^{n}p^{p}\ldots (\frac{a+b+c+\ldots}{m+n+p+\ldots})^{m+n+p+\ldots}$.

Some examples using the above techniques:

Example 1:

Show that $(1^{r}+2^{r}+3^{r}+\ldots+n^{r})>n^{n}(n!)^{r}$ where r is any real number.

Solution 1:

Since $\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n}>(1^{r}.2^{r}.3^{r}\ldots n^{r})^{\frac{1}{n}}$

Hence, $(\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n})^{n}>1^{r}.2^{r}.3^{r} \ldots n^{r}$

that is, $>(n!)^{r}$, which is the desired result.

Example 2:

Find the greatest value of $(a+x)^{3}(a-x)^{4}$ for any real value of x numerically less than a.

Solution 2:

The given expression is greatest when $(\frac{a+x}{3})^{3}(\frac{a-x}{4})^{4}$ is greatest; but, the sum of the factors of this expression is $3(\frac{a+x}{3})+4(\frac{a-x}{4})$, that is, $2a$; hence, $(a+x)^{3}(a-x)^{4}$ is greatest when $\frac{a+x}{3}=\frac{a-x}{4}$, that is, $x=-\frac{a}{7}$. Thus, the greatest value is $\frac{6^{3}8^{4}}{7^{7}}a^{r}$.

Some remarks/observations:

The determination of maximum and minimum values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:

Question:

Divide an odd integer into two integral parts whose product is a maximum.

Let an odd integer be represented as $2n+1$; the two parts by x and $2n+1-x$; and the product by y; then $(2n+1)x-x^{2}=y$; hence,

$2x=(2n+1)\pm \sqrt{(2n+1)^{2}-4y}$

but the quantity under the radical sign must be positive, and therefore y cannot be greater than $\frac{1}{4}(2n+1)^{2}$, or, $n^{2}+n+\frac{1}{4}$; and since y is integral its greatest value must be $n^{2}+n$; in which case $x=n+1$, or n; thus, the two parts are n and $n+1$.

Sometimes we may use the following method:

Find the minimum value of $\frac{(a+x)(b+x)}{c+x}$.

Solution:

Put $c+x=y$; then the expression $=\frac{(a-c+y)(b-c+y)}{y}=\frac{(a-c)(b-c)}{y}+y+a-c+b-c$

which in turn equals

$(\frac{\sqrt{(a-c)(b-c)}}{\sqrt{y}}-\sqrt{y})^{2}+a-c+b-c+2\sqrt{(a-c)(b-c)}$.

Hence, the expression is a a minimum when the square term is zero; that is when $y=\sqrt{(a-c)(b-c)}$.

Thus, the minimum value is $a-c+b-c+2\sqrt{(a-c)(b-c)}$, and the corresponding value of x is $\sqrt{(a-c)(b-c)}-c$.

Problems for Practice:

1. Find the greatest value of x in order that $7x^{2}+11$ may be greater than $x^{3}+17x$.
2. Find the minimum value of $x^{2}-12x+40$, and the maximum value of $24x-8-9x^{2}$.
3. Show that $(n!)^{2}>n^{n}$ and $2.4.6.\ldots 2n<(n+1)^{n}$.
4. Find the maximum value of $(7-x)^{4}(2+x)^{5}$ when x lies between 7 and -2.
5. Find the minimum value of $\frac{(5+x)(2+x)}{1+x}$.

More later,

Nalin Pithwa.