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Telescoping Series

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

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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 $\sqrt{2}$?

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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.