# Huygens’ problem to Leibnitz: solution

In the Feb 23 2018 blog problem, we posed the following question:

Sum the following infinite series:

$1+\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15}+ \ldots$.

Solution:

The sum can be written as:

$S=\sum_{n=1}^{\infty}P_{n}$, where $P_{n}=\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1})$.

Thus, $2(1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots)=2$. This is the answer.

If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.

Cheers,

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.

# 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.