# 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.

# Functions — “s’wat” Math is about !! :-)

Reference: Nordic Mathematical Contest 1987, R. Todev:

Question:

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions $f(2)=a>2$ and $f(mn)=f(m)f(n)$ for all natural numbers m and n. Define the smallest possible value of a.

Solution:

Since, $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that $a=3$. It is easy to prove by induction that $f(n^{k})={f(n)}^{k}$ for all $k \geq 1$. So, taking into account that f is strictly increasing, we get

${f(3)}^{4}=f(3^{4})=f(81)>f(64)=f(2^{6})={f(2)}^{6}=3^{6}=27^{2}>25^{2}=5^{4}$

as well as ${f(3)}^{8}=f(3^{8})=f(6561).

So, we arrive at $5. But, this is not possible, since $f(3)$ is an integer. So, $a=4$.

Cheers,

Nalin Pithwa.

# Famous harmonic series questions

Question 1:

The thirteenth century French polymath Nicolas Oresme proved that the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots$ does not converge. Prove this result.

Question 2:

Prove that $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots$ does not converge.

Question 3:

Prove that $1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \ldots$ does not converge.

Even if you solve these all on your own, you won’t achieve the glory of that French polymath, but you will have “re-discovered” some “elements of truth of analysis” …You can give yourself a pat on the back!

Cheers,

Nalin Pithwa.

# Infinity of primes — proof using calculus

Let us look at a proof that the number of primes is infinite — a proof that uses only elementary calculus.

Let $\pi (x) = \# \{ p \leq x: p \in P\}$ be the number of primes that are less than or equal to the real number x. (Note: here P is the set of primes). We number the primes $P = \{ p_{1}, p_{2}, p_{3, \ldots}\}$ in increasing order. Consider the natural logarithm $\log {x}$ defined as $\log {x}=\int_{1}^{x}\frac{1}{t}dt$.

Now we compare the area below the graph of $f(t)=\frac{1}{t}$ with an upper step function. Thus, for $n \leq x < n+1$ we have

$\log {x} \leq 1+\frac{1}{2}+\frac{1}{3}+ \ldots + \frac{1}{n-1}+\frac{1}{n} \leq \Sigma \frac{1}{m}$ where the sum extends over all $m \in N$ which have only prime divisors $p \leq x$.

Since every such m can be written in a unique way as a product of the form $\Pi_{p \leq x}p^{k_{p}}$, we see that the last sum is equal to

$\Pi_{{p \in P}_{p \leq x}}(\Sigma_{k \geq 0}\frac{1}{p^{k}})$

The inner sum is a geometric series with ratio $\frac{1}{p}$, hence,

$\log {x} \leq \Pi_{{p \in P}_{p \leq x}}\frac{1}{1-\frac{1}{p}}=\Pi_{{p \in P}_{p \leq x}}\frac{p}{p-1}=\Pi_{k=1}^{\pi(x)}\frac{p_{k}}{p_{k}-1}$

Now, clearly $p_{k} \geq k+1$ and thus

$\frac{p_{k}}{p_{k}-1}=1+ \frac{1}{p_{k}-1} \leq 1 + \frac{1}{k}=\frac{k+1}{k}$

and therefore,

$\log {x}\leq \Pi_{k=1}^{\pi(x)}\frac{k+1}{k}=\pi(x)+1$.

Everybody knows that $\log {x}$ is not bounded, so we conclude that $\pi(x)$ is unbounded as well, and so there are infinitely many primes. QED.

Ref: Proofs from THE BOOK (Martin Aigner and Gunter M. Ziegler) (Third Edition).

Nalin Pithwa

# Real Numbers, Sequences and Series: part 9

Definition.

We call a sequence $(a_{n})_{n=1}^{\infty}$ a Cauchy sequence if for all $\varepsilon >0$ there exists an $n_{0}$ such that $|a_{m}-a_{n}|<\varepsilon$ for all m, $n > n_{0}$.

Theorem:

Every Cauchy sequence is a bounded sequence and is convergent.

Proof.

By definition, for all $\varepsilon >0$ there is an $n_{0}$ such that

$|a_{m}-a_{n}|<\varepsilon$ for all m, $n>n_{0}$.

So, in particular, $|a_{n_{0}}-a_{n}|<\varepsilon$ for all $n > n_{0}$, that is,

$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.

Let $M=\max \{ a_{1}, \ldots, a_{n_{0}}, a_{n_{0}+1}+\varepsilon\}$ and $m=\min \{ a_{1}, \ldots, a_{n_{0}+1}-\varepsilon\}$.

It is clear that $m \leq a_{n} \leq M$, for all $n \geq 1$.

We now prove that such a sequence is convergent. Let $\overline {\lim} a_{n}=L$ and $\underline{\lim}a_{n}=l$. Since any Cauchy sequence is bounded,

$-\infty < l \leq L < \infty$.

But since $(a_{n})_{n=1}^{\infty}$ is Cauchy, for every $\varepsilon >0$ there is an $n_{0}=n_{0}(\varepsilon)$ such that

$a_{n_{0}+1}-\varepsilon for all $n>n_{0}$.

which implies that $a_{n_{0}+1}-\varepsilon \leq \underline{\lim}a_{n} =l \leq \overline{\lim}a_{n}=L \leq a_{n_{0}+1}+\varepsilon$. Thus, $L-l \leq 2\varepsilon$ for all $\varepsilon>0$. This is possible only if $L=l$.

QED.

Thus, we have established that the Cauchy criterion is both a necessary and sufficient criterion of convergence of a sequence. We state a few more results without proofs (exercises).

Theorem:

For sequences $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$.

(i) If $l \leq a_{n} \leq b_{n}$ and $\lim_{n \rightarrow \infty}b_{n}=l$, then $(a_{n})_{n=1}^{\infty}$ too is convergent and $\lim_{n \rightarrow \infty}a_{n}=l$.

(ii) If $a_{n} \leq b_{n}$, then $\overline{\lim}a_{n} \leq \overline{\lim}b_{n}$, $\underline{\lim}a_{n} \leq \underline{\lim}b_{n}$.

(iii) $\underline{\lim}(a_{n}+b_{n}) \geq \underline{\lim}a_{n}+\underline{\lim}b_{n}$

(iv) $\overline{\lim}(a_{n}+b_{n}) \leq \overline{\lim}{a_{n}}+ \overline{\lim}{b_{n}}$

(v) If $(a_{n})_{n=1}^{\infty}$ and $(b_{n})_{n=1}^{\infty}$ are both convergent, then $(a_{n}+b_{n})_{n=1}^{\infty}$, $(a_{n}-b_{n})_{n=1}^{\infty}$, and $(a_{n}b_{n})_{n=1}^{\infty}$ are convergent and we have $\lim(a_{n} \pm b_{n})=\lim{(a_{n} \pm b_{n})}=\lim{a_{n}} \pm \lim{b_{n}}$, and $\lim{a_{n}b_{n}}=\lim {a_{n}}\leq \lim {b_{n}}$.

(vi) If $(a_{n})_{n=1}^{\infty}$, $(b_{n})_{n=1}^{\infty}$ are convergent and $\lim_{n \rightarrow \infty}b_{n}=l \neq 0$, then $(\frac{a_{n}}{b_{n}})_{n=1}^{\infty}$ is convergent and $\lim_{n \rightarrow \frac{a_{n}}{b_{n}}}= \frac{\lim {a_{n}}}{\lim{b_{n}}}$.

Reference: Understanding Mathematics by Sinha, Karandikar et al. I have used this reference for all the previous articles on series and sequences.

More later,

Nalin Pithwa

# Real Numbers, Sequences and Series: part 8

Definition. A sequence is a function $f:N \rightarrow \Re$. It is usual to represent the sequence f as $(a_{n})_{n=1}^{\infty}$ where $f(n)=a_{n}$.

Definition. A sequence $(a_{n})_{n=1}^{\infty}$ is said to converge to a if for each $\varepsilon>0$ there is an $n_{0}$ such that

$|a_{n}-a|<\varepsilon$ for all $n > n_{0}$.

It can be shown that this number is unique and we write $\lim_{n \rightarrow \infty}a_{n}=a$.

Examples.

(a) The sequence $\{ 1, 1/2, 1/3, \ldots, 1/n \}$ converges to 0. For $\varepsilon>0$, let $n_{0}=[\frac{1}{\varepsilon}]+1$. This gives

$|\frac{1}{n}-0|=\frac{1}{n}<\varepsilon$ for all $n>n_{0}$.

(b) The sequence $\{ 1, 3/2, 7/4, 15/8, 31/16, \ldots\}$ converges to 2. The nth term of this sequence is $\frac{2^{n}-1}{2^{n-1}}=2-\frac{1}{2^{n-1}}$. So $|2-(2-\frac{1}{2^{n-1}})|=\frac{1}{2^{n-1}}$. But, $2^{n-1} \geq n$ for all $n \geq 1$. Thus, for a given $\varepsilon >0$, the choice of $n_{0}$ given in (a) above will do.

(c) The sequence $(a_{n})_{n=1}^{\infty}$ defined by $a_{n}=n^{\frac{1}{n}}$ converges to 1.

Let $n^{\frac{1}{n}}=1+\delta_{n}$ so that for $n >1$, $\delta_{n}>0$. Now, $n=(1+\delta_{n})^{n}=1+n\delta_{n}+\frac{n(n-1)}{2}(\delta_{n})^{2}+\ldots \geq 1+\frac{n(n-1)}{2}(\delta_{n})^{2}$, thus, for $n-1>0$, we have $\delta_{n} \leq \sqrt{\frac{2}{n}}$. For any $\varepsilon > 0$, we can find $n_{0}$ in N such that $n_{0}\frac{(\varepsilon)^{2}}{2}>1$. Thus, for any $n > n_{0}$, we have $0 \leq \delta_{n} \leq \sqrt{\frac{2}{n}} \leq \sqrt{\frac{2}{n_{0}}}<\varepsilon$. This is the same as writing

$|n^{\frac{1}{n}}-1|<\varepsilon$ for $n > n_{0}$ or equivalently, $\lim_{n \rightarrow \infty}n^{\frac{1}{n}}=1$

(d) Let $a_{n}=\frac{2^{n}}{n!}$. The sequence $(a_{n})_{n=1}^{\infty}$ converges to o. Note that

$a_{n}=\frac{2}{1}\frac{2}{2}\frac{2}{3}\ldots\frac{2}{n}<\frac{4}{n}$ for $n>3$.

For $\varepsilon>0$, choose $n_{1}$ such that $n_{1}\varepsilon>4$. Now, let $n_{0}=max \{ 3, n\}$ so that $|a_{n}|<\varepsilon$ for all $n > n_{0}$.

(e) For $a_{n}=\frac{n!}{n^{n}}$, the sequence $(a_{n})_{n=1}^{\infty}$ converges to 0. (Exercise!)

In all the above examples, we somehow guessed in advance what a sequence converges to. But, suppose we are not able to do that and one asks whether it is possible to decide if the sequence converges to some real number. This can be helped by the following analogy: Suppose there are many people coming to Delhi to attend a conference. They might be taking different routes. But, as soon as they come closer and closer to Delhi the distance between the participants is getting smaller.

This can be paraphrased in mathematical language as:

Theorem. If $(a_{n})_{n=1}^{\infty}$ is a convergent sequence, then for every $\varepsilon>0$, we can find an $n_{0}$ such that

$|a_{n}-a_{m}|<\varepsilon$ for all $m, n > n_{0}$.

Proof. Suppose $(a_{n})_{n=1}^{\infty}$ converges to a. Then, for every $\varepsilon$ we can find an $n_{0}$ such that

$|a_{n}-a|<\varepsilon/2$ for all $n > n_{0}$.

So, for $m, n > n_{0}$, we have

$|a_{m}-a_{n}|=|a_{n}-a+a-a_{m}|\leq |a_{n}-a|+|a-a_{m}|<\varepsilon$. QED.

The proof is rather simple. This very useful idea was conceived by Cauchy and the above theorem is called Cauchy criterion for convergence. What we have proved above tells us that the criterion is a necessary condition for convergence. Is it also sufficient? That is, given a sequence $(a_{n})_{n=1}^{\infty}$ which satisfies the Cauchy criterion, can we assert that there is a real number to which it converges? The answer is yes!

We call a sequence $(a_{n})_{n=1}^{\infty}$ monotonically non-decreasing if $a_{n+1} \geq a_{n}$ for all n.

Theorem. A monotonically non-decreasing sequence which is bounded above converges.

Proof. Suppose $(a_{n})_{n=1}^{\infty}$ is monotonic and non-decreasing. We have $a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq a_{n+1} \leq \ldots$. Since the sequence is bounded above, $\{ a_{k}: k=1, \ldots\}$ has a least upper bound. Let it be a. By definition, $a_{n} \leq a$ for all n, but for $\varepsilon>0$ there is at least one $n_{0}$ such that $a_{n_{0}}+\varepsilon>a$. Therefore, for $n > n_{0}$, $a_{n}+\varepsilon \geq a_{n_{0}}+\varepsilon>a\geq a_{n}$. This gives us $|a_{n}-a|<\varepsilon$ for all $n \geq n_{0}$. QED.

We can similarly prove that: A monotonically non-increasing sequence which is bounded below is convergent.

Suppose we did not have the condition of boundedness below or above for a monotonically non-increasing or non-decreasing sequence respectively, then what would happen? If a sequence is monotonically non-decreasing and is not bounded above, then given any real number $M>0$ there exists at least one $n_{0}$ such that $a_{n_{0}}>M$, and hence $a_{n}>M$ for all $n > n_{0}$. In such a case, we say that $(a_{n})_{n=1}^{\infty}$ diverges to $\infty$. We write $\lim_{n \rightarrow \infty}a_{n}=\infty$. More generally, (that is, even when the sequence is not monotone), the same criterion above allows us to say that $\{ a_{n}\}_{n=1}^{\infty}$ diverges to $\infty$ and we write $\lim_{n \rightarrow \infty}a_{n}=\infty$. We can similarly define divergence to $-\infty$.

We make a digression here: In case a set is bounded above, then we have the concept of least upper bound. For any set A or real numbers, we define supremum of A as

$sup \{ x: x \in A\}=sup A = least \hspace{0.1in} upper \hspace{0.1in}bound \hspace{0.1in} of \hspace{0.1in} A$, if A is bounded above and is equal to $\infty$ if A is not bounded above.

Similarly, we define infimum of a set A as

$inf \{ x: x \in A\}= inf A= greatest \hspace{0.1in} lower \hspace{0.1in} bound \hspace{0.1in} of \hspace{0.1in}A$, if A is bounded below, and is equal to $-\infty$, if A is not bounded below.

This would help us to look for other criteria for convergence. For any bounded sequence (a_{n})_{n=1}^{\infty} of real numbers, let

$b_{n} = \sup \{ a_{n}, a_{n+1}, a_{n+2}, \ldots\}=\sup \{ a_{k}: k \geq n\}$.

It is clear that $(b_{n})_{n=1}^{\infty}$ is now a non-increasing sequence. So, it converges. We set

$\lim_{n \rightarrow \infty}b_{n}=limit \hspace{0.1in} superior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \sup a_{n}= \overline{\lim}_{n}a_{n}$,

Similarly, if we write

$a_{n}=\inf \{ a_{n}, a_{n+1}, \ldots\}=\inf \{ a_{k}: k \geq n\}$,

then $(a_{n})_{n=1}^{\infty}$ is a monotonically non-decreasing sequence. So we write

$\lim_{n \rightarrow \infty}c_{n}=limit \hspace{0.1in} inferior \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} sequence \hspace{0.1in} (a_{n})_{n=1}^{\infty}= \lim \inf a_{n}= \underline{\lim}_{n}a_{n}$,

We may not know a sequence to be convergent or divergent, yet we can find its limit superior and limit inferior.

We have in fact the following result:

Theorem.

For a sequence $(a_{n})_{n=1}^{\infty}$ of real numbers, $\underline{\lim}a_{n} \leq \overline{\lim}a_{n}$.

Further, if $l=\underline{\lim}a_{n}$, and $L=\overline{\lim}a_{n}$ are finite, then $l=L$ if and only if the sequence is convergent.

Proof.

It is easy to see that $l \leq L$. Now, suppose l, L are finite.If $l=L$ are finite. If $l=L$, then for all $\varepsilon >0$ there exists $n_{1}, n_{2}$ such that

$l - \varepsilon< \sup_{k \geq n} a_{k}< l + \varepsilon$ for all $n \geq n_{1}$ and $l - \varepsilon < \inf_{k \geq n} < l + \varepsilon$ for all $n > n_{2}$.

Thus, with $n_{0}=max{n_{1},n_{2}}$ we have

$l - \varepsilon for all $n > n_{0}$.

This proves that equality holds only when it is convergent. Conversely, suppose $(a_{n})_{n=1}^{\infty}$ converges to a. For every $\varepsilon > 0$, we have $n_{0}$ such that $a-\varepsilon for all $n > n_{0}$. Therefore, $\sup_{k \geq n}a_{k} \leq a+\varepsilon$ and

$a-\varepsilon \leq \inf a_{n}$ for all $n \geq n_{0}$. Hence, $a-\varepsilon \leq < L \leq a+\varepsilon$. Since this is true for every $\varepsilon > 0$, we have $L=l=a$.

QED.

# Real Numbers, Sequences and Series: Part 6

Theorem:

Given $x \in \Re_{+}$ and $n \in N$, we can find $y \in \Re$ such that $x=y^{n}$.

Proof:

Let $A={u \in \Re_{+}: u^{n}. If $x<1$, then $x^{n} and hence $x \in A$. On the other hand, if $x \geq 1$, then $1/2 \in A$. Thus, A is non-empty. Next, observe that if $v^{n}>x$, then v is an upper bound of A. In particular, $\{ 1+x\}$ is an upper bound of A.

By the least upper bound property, A admits a least upper bound. Let us denote it by y. We will rule out the possibilities $y^{n}>x$ and $y^{n} implying that $x=y^{n}$.

If $y^{n}, let $a=\frac{x-y^{n}}{nx}$ and $z=y(1+a)$. It can be checked that $z^{n} so that $z \in A$. But, $y , contradicting the fact that y is the least upper bound of A. (we have used the inequalities 7 and 8 in the previous blog).

On the other hand, if $y^{n}>x$, let $\frac{y^{n}-x}{ny^{n}}$ and $w=y(1-b)$. Again, it can be verified that $w^{n}>x$ and hence w is an upper bound of A. But, $w, contradicting the fact that y is the least upper bound of A. Hence, $y^{n}=x$QED.

In particular, we see that there is an element $\alpha \in \Re$ such that $\alpha^{2}=2$ and hence also $(-\alpha)^{2}=2$ which means that the equation $x^{2}=2$ has two solutions. The positive one of those two solutions is $\sqrt{2}$. In fact, the above theorem has guaranteed its extraction of the square root, cube root, even nth root of any positive number. You could ask at this stage, if this guarantees us extraction of square root of a negative number. The answer is clearly no. Indeed, we have

$x^{2} \geq 0$ for $x \in \Re$.

Remark.

We can further extend $\Re$ to include numbers whose squares are negative, which you know leads to the idea of complex numbers.

We have shown that Q is a subset of $\Re$. One can also show that between any two distinct real numbers there is a rational number. This naturally leads to the decimal representation of real number: Given any real number x and any $q \in N$, we can get a unique $a_{0} \in Z$ and unique $a_{1},a_{2}, \ldots a_{q} \in N$ such that $0 \leq a_{1} \leq a_{2} \ldots a_{q} \leq 9$ and

$|x-(a_{0}+a_{1}/10+a_{2}/100+\ldots + \frac{a_{q}}{10^{q}})|<\frac{1}{10^{q}}$

You are invited to try to  prove this familiar decimal representation.

If we have a terminating decimal representation of a real number, then surely it is rational.But, we know that rationals like 1/3, 1/7, 1/11, do not have a terminating decimal expansion.

It is clear that the decimal representation of $\sqrt{2}$ cannot terminate as it is not rational. There are many elements of $\Re$ which are not in Q. Any element of $\Re$ which is not in is called an irrational number, and irrational numbers cannot have terminating decimal representation.

More later,

Nalin Pithwa