# Thread: Size!

1. ## Size!

Problem: Let $n\in\mathbb{N}$. Which is larger $99^n+100^n$ or $101^n$? (there may be more than one case)

Extra: Can a generlization be made about the bases?

2. Ok this hasn't been thought out well but I thought I would write down my initial thoughts.
For an n you have 1 set of (99 + 100) and 1 set of 101 on its own (I can't explian whats going on in my head for this part!).
The difference between the two is 98 so my guess is the change from $99^n + 100^n > 101^n$ to $99^n + 100^n < 101^n$ would happen at n=49.
Why I cant fully explain but something along the lines of that there's two things n acts on in on the LHS and one on the RHS and the RHS will 'catch' up to the LHS by 2 each time........

Ill think some more about what I'm actually trying to say but for now I think the change happens at n=49.

3. Umm... To further my ramblings...

Want to check if $99^n + 100^n - 101^n < 0$.

i.e. that $99^n + 100^n - 101^n = 100^n(0.99^n + 1^n - 1.01^n) = 100^n(0.99^n + 1 - 1.01^n) < 0$
$=> 1 + 0.99^n < 1.01^n$.

Clearly $1 + 0.99^n > 1$ and $1.01^n > 1$, $1 + 0.99^n$ is converging towards 1 while $1.01^n$ is monotonically increasing.

Hence at some point $1.01^n$ will exceed $1 + 0.99^n$ for some value of $n$. Just remains to find this $n$ as up until this point $99^n + 100^n > 101^n$.

Pretty sure this is 49 but I don't know how to provide a proper proof!

4. Originally Posted by Deadstar
Umm... To further my ramblings...

Want to check if $99^n + 100^n - 101^n < 0$.

i.e. that $99^n + 100^n - 101^n = 100^n(0.99^n + 1^n - 1.01^n) = 100^n(0.99^n + 1 - 1.01^n) < 0$
$=> 1 + 0.99^n < 1.01^n$.

Clearly $1 + 0.99^n > 1$ and $1.01^n > 1$, $1 + 0.99^n$ is converging towards 1 while $1.01^n$ is monotonically increasing.

Hence at some point $1.01^n$ will exceed $1 + 0.99^n$ for some value of $n$. Just remains to find this $n$ as up until this point $99^n + 100^n > 101^n$.

Pretty sure this is 49 but I don't know how to provide a proper proof!
Spoiler:
It is indeed true if $n\ge 49$! Now try to find a nice proof for it.

5. Partial solution

Spoiler:
The above inequality is true when $n\ge 49$

Proof: This is equivalent to showing that $\frac{101^n=99^n}{100^n}>1$. But $101^n-99^n=\left(100+1\right)^n-\left(100-1\right)^n=\sum_{\ell=1}^{n} {n \choose \ell}100^{n-\ell}$ $-\sum_{\ell=0}^{n} {n\choose \ell}100^{n-\ell}(-1)^{\ell}=\sum_{n=0}^{n}{n\choose \ell}100^{n-\ell}\left(1-\left(-1\right)^{\ell}\right)\color{red}\star$. Now clearly $1-(-1)^{\ell}=\begin{cases} 0 & \mbox{if}\quad \ell\text{ is even} \\ 1 & \mbox{if}\quad \ell\text{ is odd} \end{cases}$. So $\color{red}\star$ reduces to $2\sum_{\ell=0}^{\frac{n-1}{2}}{n\choose {2\ell-1}}100^{n-2\ell+1}=2\left(100^{n-1}n+\frac{n(n-1)(n-2)100^{n-3}}{ 6}+\cdots\right)$. Therefore $\frac{101^n-99^n}{100^n}=2\left(\frac{n}{100}+\frac{n(n-1)(n-2)}{100^3\cdot 6}+\cdots\right)$. If $n\ge50$ this is trivial since $\frac{101^n-99^n}{100^n}>2\cdot\frac{n}{100}>1$. For the case where $n=49$ merely note that $2\cdot\left(\frac{49}{100}+\frac{49\cdot48\cdot47} {100^3\cdot 6}+\cdots\right)>2\cdot\left(\frac{49}{100}+\frac{ 49\cdot48\cdot47}{100^3\cdot 6}\right)>2\left(\frac{49}{100}+\frac{1}{100}\righ t)>1$