# Simple proof that $8\left(\frac{9}{10}\right)^8 > 1$

This question is motivated by a step in the proof given here.

\begin{align*} 8^{n+1}-1&\gt 8(8^n-1)\gt 8n^8\\ &=(n+1)^8\left(8\left(\frac{n}{n+1}\right)^8\right)\\ &\geq (n+1)^8\left(8\left(\frac{9}{10}\right)^8\right)\\ &\gt (n+1)^8 . \end{align*}

I had no trouble following along with the proof until I hit the step that relied on $$8\left(\frac{9}{10}\right)^8 > 1$$. So I whipped out a calculator and confirmed that this is indeed correct. And I could see, after some fooling around with pen and paper that any function in the form \begin{align} k \; \left(\frac{n}{n+1}\right)^k \end{align} where $n \in \mathbb{Z}$ and $k \rightarrow \infty$ is bound to fall below one and stay there. So it's not a given that any function in the above form will be greater than one.

What I'm actually curious about is whether there are nifty or simple little tricks or calculations you can do in your head or any handwavy type arguments that you can make to confirm that $$8\left(\frac{9}{10}\right)^8 > 1$$ and even more generally, to confirm for certain natural numbers $k,n$ whether \begin{align} k \; \left(\frac{n}{n+1}\right)^k > 1 \end{align}

So are there? And if there are, what are they?

It can be geometrical. It can use arguments based on loose bounds of certain log values. It doesn't even have to be particularly simple as long as it is something you can do in your head and it is something you can explain reasonably clearly so that others can do also it (so if you're able to mentally calculate numbers like Euler, it's not useful for me).

You can safely assume that I have difficulties multiplying anything greater two single digit integers in my head. But I do also know that $$\limsup_{k\rightarrow \infty} \log(k) - a\cdot k < 0$$ for any $a>0$ without having to go back to a textbook.

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Bernoulli's Inequality might be what you're looking for. –  lvb Mar 6 '11 at 18:48
For the general case, you can take logs to $\ln k + k\ln(1-\frac{1}{n+1})\gt 0$ then use the expansion of ln (for large $n$) to get $\ln k \gt \frac{k}{n+1}$ or $n \gt \frac{k}{\ln k}-1$ which works well if you have some logs in your head (2, 10?) –  Ross Millikan Mar 7 '11 at 3:27

Note that $(1-.1)^8\geq 1-.8$ by Bernoulli's inequality, as mentioned in a comment by lvb. For other cases you can use $\left(1-\frac{1}{n+1}\right)^k\geq 1-\frac{k}{n+1}$, which makes it easier to find a sufficient condition on $n$ for $k\left(\frac{n}{n+1}\right)^k$ to be larger than $1$.

In the motivating problem, the idea was that $k$ is fixed (at $8$), and since $\frac{n}{n+1}$ goes to $1$ as $n$ goes to infinity, $k\left(\frac{n}{n+1}\right)^k$ approaches $k$, and in particular is eventually bigger than $1$. One then finds out in a particular case what "eventually" means; here $n=9$ suffices. In general, Bernoulli's inequality leads to the conclusion that $n>k$ suffices.

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Thanks! Chosen for answering the general case as well. –  JasonMond Mar 6 '11 at 21:38
(+1) Good to spell out the general case. –  cardinal Mar 6 '11 at 23:12

For this particular example, it's pretty easy, since $$(9/10)^8 \geq (8/10)^4 \geq (6/10)^2 \geq (3/10)$$ and so you're done. Notice, my bounds are quite crude in the last two cases.

Repetitive squaring is the quickest way to get up to a large power. In fact, that's how a typical computer implementation will do it for integer powers.

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This is nice; it uses the fact that the denominator is 10, so the approximations amount to squaring integers and keeping only the first digit. –  mjqxxxx Mar 6 '11 at 18:57
(+1) for the same reason as mjqxxxx explains. Really nice! –  Gottfried Helms Mar 6 '11 at 19:33
Ditto on the +1. I'd never heard of repetitive squaring before. Cool. –  JasonMond Mar 6 '11 at 21:39

You can use the fact that $(1 + x/n)^n$ approaches $e^x$ for large $n$. Then $$\left(\frac{n-1}{n}\right)^{k} = \left(1 - \frac{1}{n}\right)^{k} \thicksim e^{-k/n}$$ as $n\rightarrow\infty$. In your case, this would give $(9/10)^8 \approx e^{-4/5} > 1/e$, which is clearly greater than $1/8$. (For $n=10$ and $x=-1$, the error in the exponential approximation is about 5%.)

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Thanks! I can also imagine some pairs of $(k,n)$ where I'd run into trouble establishing an inequality in my head. –  JasonMond Mar 6 '11 at 21:43
(+1) And, via this heuristic, one can get rough upper and lower "bounds" via $1 - (k/n) \leq e^{-k/n} \leq 1 - (k/n) + (k/n)^2$. Note that the LHS (heuristically) "recovers" Bernoulli's inequality. Of course, since, in reality, $(1-1/n)^k \leq e^{-k/n}$, my statements are all quite crude. –  cardinal Mar 6 '11 at 23:11

I'll start from the "obvious" fact that $(5/4)^3 < 2$. In fact, the cube root of 2 is around 1.26; of course you can explicitly compute $(5/4)^3 = 125/64 < 128/64 = 2$.

Then cubing both sides of that inequality, you get

$(5/4)^9 < 8$.

But $(10/9)^8 < (10/9)^9 < (5/4)^9$, so $(10/9)^8 < 8$. Taking reciprocals, $(9/10)^8 > 1/8$; multiplying both sides by 8 gives your result.

Actually, my heuristic here is as follows: $(10/9)$ is roughly one whole tone (two semitones); so $(10/9)^8$ is around sixteen semitones, or an octave and a major third, or $2 \times (5/4) = 2.5$. So $(9/10)^8$ must be around $0.4$. See Sanjoy Mahajan's handout on "singing logarithms", originally due to I. J. Good. Of course this method is really only useful if you know a little music theory.

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Ok, here is another one:

$$8\left(\frac{9}{10}\right)^8 > 8\left(\frac{9}{10}\right)^9= \left[ 2\left(\frac{9}{10}\right)^3 \right]^3 \,.$$

So if we can prove

$$2\left(\frac{9}{10}\right)^3 >1 \,,$$

we are done.

This is equivalent to

$$3^6 > 2^25^3$$

Which is true since

$$3^3 > 5^2$$ and $$3^3 > 2^25$$

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Just another mental approach. We write $$8\left(\frac9{10}\right)^8 >1 \rightarrow \frac9{10}>\frac18^{\frac18}=\left(1-\frac78\right)^{\frac18}$$ Then if one remembers that $(1-x)^{\frac1a} = 1 - \frac1a x - O\left(\frac{x^2}a\right)$ then we get $$\frac9{10}> 1-\frac7{64}-\epsilon \rightarrow 9>10-\frac{70}{64}-10\epsilon$$ which is obviously true. It just depends on whether you remember the beginning of the binomial-formula for fractional exponents.

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Here is one that uses two facts about powers of 2 and 10: $2^{10}=1024>10^3$ and $2^7=128>10^2$

$$8\left(\frac{9}{10}\right)^8>8\left(\frac{8}{10}\right)^8=\frac{8^9}{10^8}=\frac{2^{3·9}}{10^6·10^2}>\frac{2^{27}}{2^{20}·10^2}=\frac{2^7}{10^2}=\frac{128}{100}>1$$

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