### Telescoping Series

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

### Growing

Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)

### 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 $\sqrt{2}$?

# Big, Bigger, Biggest

##### Stage: 5 Challenge Level:

Which is the biggest and which the smallest of these numbers and how do they compare in magnitude?

$$A = 2000^{2002},\ B = 2001^{2001},\ C = 2002^{2000}$$

This solution comes from Ilham, St. Patrick's College, Wellington, well done and thank you Ilham.

First let's define the function floor($x$), where $x$ is a real number, such that floor($x$) = the integer part of $x$.

Let $$y = \rm{floor}(\log_a (x)) + 1$$.

As a general rule, y will be the number of digits of $x$ in base $a$. If we reverse this, we can say that $x$ is somewhere between $a ^ y$ and $a^{y + 1}$.

Another basic rule is $\log_a (b^c) = c\log_a (b)$. If we don't use this rule, the calculation cannot be handled using any standard scientific calculators, as they can't handle calculation with numbers greater than $10^{100}$.

If we use these two rules to $A$, $B$ and $C$ in base $10$, it will show that $A$ has $6609$ digits, $B$ has $6606$ digits, and $C$ has $6603$ digits in base 10.

Therefore, $A$ is bigger than $B$ which in turn is bigger than $C$. $A$ is the biggest, and $C$ is the smallest.

A similar solution uses the fact that the logarithm function is an increasing function so it follows that $$\log A > \log B$$ if and only if $A > B$. Hence

$$\log A = 2002 \log 2000 \approx 2002(3.010) \approx 6608.662$$ $$\log B = 2001 \log 2001 \approx 6605.795$$ $$\log C = 2000 \log 2002 \approx 6602.928$$

The approximate difference is given by : $\log A - \log B = \log A/B \approx 3$, hence $A\approx 10^3B$. Similarly $B\approx 10^3C$. Thus $A > B > C.$

Here is Koopa Koo's more general result.

Claim: $A > B > C$

Proof: $A > B$ if and only if $\log A > \log B.$

I shall prove $\log A - \log B > 0$ i.e. $2002\log2000 - 2001\log2001 > 0.$

Let $f(x) = (x + 2)\log x - (x+1)\log(x+1)$ so that for example f(2) = 4log2 - 3log3.

Differentiating this function, $$f'(x) = (x + 2)/x + \log x - 1 - \log(x + 1) = 2/x -\log[(x+1)/x].$$

This derivative is positive if and only if $e^{2/x}> (x+1)/x.$

Using $e^y > 1 + y$ for all $y$, let $y = 2/x$.

We have $e^{2/x}> 1 + 2/x = (x + 2)/x > (x + 1)/x$.

So the function f is increasing, in particular, $f(2000) = \log A - \log B > 0$ and it follows that $A > B$.

The proof that $B > C$ is similar.