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There was a problem in Problem-Solving Strategies. I have a quite similar one, but here we need to find such a number, not prove its existence...

Find a positive integer $n>1$ satisfying $\sqrt[n]{n}<1+10^{-20}$.

Again, the problem is not to prove that there exist one, I need a method to find an exact number. Please help!

Bernard
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user173628192
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  • There's probably a slicker way to do this, but here is an idea that should work. Write $n = 1+x$, and expand $\sqrt[n]{n} = \exp( \frac{1}{1+x} \log(1+x) )$ as a power series in $x$. Then use the Taylor remainder theorem from calculus. – Cardboard Box Jan 17 '20 at 22:31
  • @CardboardBox, I am not sure you could get an integer value using the proposed approach. – NoChance Jan 17 '20 at 23:43

2 Answers2

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In order for $\sqrt[n]{n}<1+10^{-20}$ it suffices to find $n$ large enough such that $$ n<(1+10^{-20})^n. $$ Keeping only the third term in the binomial expansion yields $$ (1+10^{-20})^n> \binom{n}{2}10^{-40}, $$ and the right side is at least $n$ whenever $$ \frac{n-1}{2}\geq 10^{40}, $$ i.e. $n\geq 2\cdot 10^{40}+1$.

pre-kidney
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(Roughly following Arthur's proof here.)

We want to find $n$ with $n^{1/n}<1+10^{-20}$, or equivalently $(1+10^{-20})^n > n$.

First, by Bernoulli's inequality $(1+x)^r>1+rx$, we have

$$(1+10^{-20})^m > 2$$

where $m=10^{20}$. So, for any positive integer $k$, we have:

$$ (1+10^{-20})^{km} > 2^k \, . $$

If we choose $k$ such that $2^k>km$, then the desired inequality will hold for $n=km$. Since $2^7 > 10^2$, we have $2^{77}>10^{22}$, and so

$$ \frac{2^{77}}{77} > \frac{2^{77}}{100} > \frac{10^{22}}{100} > 10^{20} = m \, . $$

That is, $k=77$ will suffice and so we can take $n=77\times 10^{20}$.

Micah
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