You may also like

Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

How Many Solutions?

Find all the solutions to the this equation.


Age 16 to 18
Challenge Level

Congratulations to Federico Poloni from Casirate d'Adda (Italy) for the following solution.

Prove that $(1 + \frac{1}{n})^n \leq e < 3$.

Using Newton's formula (also called the Binomial Theorem) $$\eqalign{ \left(1 + \frac {1}{n}\right)^n &=& 1 + n\left(\frac{1}{n}\right) + \frac{n(n-1)}{2!}\left(\frac{1}{n^2}\right) + \frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n^3}\right)+ \cdots \\ &\leq & 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &\leq & e < 3. }$$ (a) Which is larger: $1.000001^{1000000}$ or $2$?

It's possible to prove that every number in the form $(1+\frac{1}{a})^a$ is greater than $2$. \[\left(1+\frac{1}{a}\right)^a = 1+a\cdot \frac{1}{a}+\mbox{other positive terms}> 2\] for every integer $a> 2$. For $a=1000000$, the problem is solved.

(b) Which is larger: $100^{300}$or $300!$ (i.e. $300$ factorial)?

This is a bit more complex. I'll use the formula $(1+ \frac{1}{n})^n = (\frac{n+1}{n})^n < 3 \quad (1)$

I will now use the last inequality to prove by induction that $ n!> (\frac{n}{3})^n.$ Clearly this is true for $n=1$ and $2$ and so using the induction hypothesis that it is true for $n$:

$(n+1)! = (n+1)n! \leq(n+1)\left(\frac{n}{3}\right)^n$

$= 3\left(\frac{n+1}{3}\right)^{n+1} \left(\frac{n}{n + 1}\right)^n$.

So using (1) gives: $$ (n+1)! \leq \left(\frac{n+1}{3}\right)^{n+1}.$$ The demonstration by induction is complete. In particular, for $n=300$, the formula solves the given problem.