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Question:

Find the smallest positive integer $n$ such that $$99^n+100^n<101^n$$

My idea: This is equivalent to $$\left(\dfrac{99}{101}\right)^n+\left(\dfrac{100}{101}\right)^n<1$$ so $$\left(1-\dfrac{2}{101}\right)^n+\left(1-\dfrac{1}{101}\right)^n<1$$

Use this: $$(1+x)^n\ge 1+nx,x>-1,n\ge 1$$ so $$1-\dfrac{2}{101}n+1-\dfrac{1}{101}n<1\Longrightarrow n>\dfrac{101}{3}=33.666$$

But I found $n=34$ is not sufficient. I used a computer to find $n\ge 49$.

I want to see how to find it by hand. Thank you.

Micah
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math110
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3 Answers3

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First of all, $n \ge 34$ is a necessary (but not sufficient) condition for that inequality to be true. Remember that $1+nx \le c$ does not imply that $(1+x)^n \le c$.

We can get a better bound by manipulating the inequality as follows:

$99^n + 100^n < 101^n$

$101^n - 99^n > 100^n$

$\left(\dfrac{101}{100}\right)^n - \left(\dfrac{99}{100}\right)^n > 1$

$\left(1+\dfrac{1}{100}\right)^n - \left(1-\dfrac{1}{100}\right)^n > 1$

Using the binomial theorem, this becomes:

$2\dbinom{n}{1}\dfrac{1}{100} + 2\dbinom{n}{3}\dfrac{1}{100^3} + 2\dbinom{n}{5}\dfrac{1}{100^5} + \cdots > 1$

All the terms on the left side are positive and first term is $\dfrac{n}{50}$, so the inequality holds for $n \ge 50$.

This is a sufficient condition, but not a necessary one. For $n = 49$, the first two terms exceed $1$, so the inequality holds for $n = 49$ as well.

Now, all that remains is to show that the inequality does not hold for $n \le 48$.

JimmyK4542
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  • I have prove $n\ge 33$,so Now the kay prove $34<n<48$ – math110 Aug 10 '14 at 03:35
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    Your expression is bounded above by $2 \sinh (n/100)$, which is roughly $0.997$ when $n=48$. So can we show that $\sinh(12/25) < 1/2$ by hand? – Micah Aug 10 '14 at 03:40
  • Showing that $\sinh(12/25) < 1/2$ is equivalant to showing that $12/25 < \sinh^{-1}(1/2) = \ln \tfrac{1+\sqrt{5}}{2}$. I don't know if that helps. – JimmyK4542 Aug 10 '14 at 04:49
  • We can come tantalizingly close by writing $\sinh^{-1}(1/2)=\int_{0}^{1/2} \frac{dx}{\sqrt{1+x^2}} > \int_0^{1/2} 1-x^2/2 , dx = 23/48$. Maybe there's some way to fix it up a little bit? – Micah Aug 10 '14 at 07:00
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Considering the problem from a purely algebraic point of view : if you plot the function $y=\log(99^n+100^n)-\log(101^n)$ (which is basically a straight line since $\frac {d^2y}{dn^2}$ is $0.0000252523$ for $n=0$ and $0.0000198134$ for $n=100$, you would see that $y=0$ if $n=48.2275$.

So the inequality holds for $n \geq 49$.

If you perform one iteration of Halley method starting at $n=0$, it predicts an underestimate of the solution at $n=48.0917$.

Claude Leibovici
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In view of JimmyK4542's answer it remains to show that $$\left({99\over101}\right)^{48}+\left({100\over101}\right)^{48}>1\ ,$$ which is the same thing as $$\left(1+{2\over99}\right)^{-48}+\left(1+{1\over100}\right)^{-48}>1\ .$$ Now $$(1+x)^{-n}=\sum_{k=0}^\infty{n+k-1\choose k}(-x)^k\ .$$ When $0<x<{k+1\over n+k}$ the terms of this alternating series are decreasing in absolute value after the $k$th term. In the following $x<{1\over 49}$, $n=48$, and $k=5$; therefore we are on the save side when we write $$(1+x)^{-48}>1 -48x+ 1176x^2 -19600x^3+ 249900x^4-2598960x^5\qquad(0<x<{1\over 49})\ .$$ Putting $x:={2\over99}$ and $x={1\over100}$ here produces $$\left(1+{2\over99}\right)^{-48}+\left(1+{1\over100}\right)^{-48}>{12404629614100147\over12382682941406250}>1\ .$$

Christian Blatter
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