Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus.

Question

Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. This inequality is equivalent to $$ 2018^{1/2018}>2019^{1/2019} $$

One of my 'High school' student asked why the inequality is true. The whole class became interested in the problem.The demonstration that such inequality is true, using calculus, can be found here. But my students are not familiar with calculus.

I can also show by induction and Newton's binomial formula that $ n^{(n + 1)}> (n + 1)^n $, to $ n> 3$, but my students are not familiar with mathematical induction. Another limitation of my students is that they have not yet learned the Newton's binomial formula.

How to prove the inequality $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without calculus? That is how to prove this inequality for High school students without using Newton's binomial formula?

Answer 1

Compared to my first answer, this answer makes a simpler use of the same familiar identity, which is proved without explicit use of induction, thus: \begin{align*} x^m - y^m & = x^m - (x^{m-1}y - x^{m-1}y) - (x^{m-2}y^2 - x^{m-2}y^2) - \cdots - (xy^{m-1} - xy^{m-1}) - y^m \\ & = (x^m - x^{m-1}y) + (x^{m-1}y - x^{m-2}y^2) + (x^{m-2}y^2 - x^{m-3}y^3) + \cdots + (xy^{m-1} - y^m) \\ & = (x - y)x^{m-1} + (x - y)x^{m-2}y + (x - y)x^{m-3}y^2 + \cdots + (x - y)y^{m-1} \\ & = (x - y)(x^{m-1} + x^{m-2}y + x^{m-3}y^2 + \cdots + y^{m-1}). \end{align*} The rest of the proof is a straightforward calculation. (It is based on a trick which was inspired loosely by nbarto's answer, but I am to blame for it!)

Suppose $n \geqslant 3$.

In the above identity, take $x = m = n+1$, $y = n$, and group together all but the first two terms in the brackets, obtaining: $$ (n+1)^{n+1} - n^{n+1} = (n+1)^n + (n+1)^{n-1}n + [(n+1)^{n-2}n^2 + \cdots + n^n]. $$ There are $n-1$ terms in the square brackets, they are in strictly decreasing order, and the largest of them is $(n+1)^{n-2}n^2$. Therefore: \begin{align*} (n+1)^{n+1} - n^{n+1} & < (n+1)^n + (n+1)^{n-1}n + (n-1)(n+1)^{n-2}n^2 \\ & = (n+1)^n + (n+1)^{n-2}[n(n+1) + n^2(n-1)] \\ & = (n+1)^n + (n^3+n)(n+1)^{n-2}. \end{align*} On the other hand: \begin{align*} (n+1)^{n+1} - (n+1)^n & = n(n+1)^n \\ & = (n+1)^n + (n-1)(n+1)^n \\ & = (n+1)^n + (n-1)(n+1)^2(n+1)^{n-2} \\ & = (n+1)^n + (n^2-1)(n+1)(n+1)^{n-2} \\ & = (n+1)^n + (n^3 + n^2 - n - 1)(n+1)^{n-2}. \end{align*} But: $$ (n^3 + n^2 - n - 1) - (n^3 + n) = n^2 - 2n - 1 = (n - 1)^2 - 2 > 0, $$ therefore: $$ (n+1)^{n+1} - n^{n+1} < (n+1)^{n+1} - (n+1)^n, $$ therefore $n^{n+1} > (n+1)^n$. $\square$

Answer 2

I would try to motivate this by showing that $f(x) = x^{1/x}$ is a monotone function for reasonably small $x$.

Another approach is to note that $$ \frac{2019^{2018}}{2018^{2018}} = \left(1 + \frac{1}{2018}\right)^{2018} $$ and so your inequality is equivalent to showing $$ \left(1 + \frac{1}{2018}\right)^{2018} < 2018, $$ which does not sound very far-fetched, since LHS is close to $e$...

UPDATE

Please see saulspatz's answer for how to prove this last claim with a hand computation only.

Answer 3

$$1+{1\over2018}<1+{1\over2000}=1.0005$$ $$\left(1+{1\over2018}\right)^{2018}<(1.0005)^{2048}$$

The right hand side can be evaluated by repeated squaring, and, since there is plenty of leeway, you can simplify the calculations by rounding up as you go. Since $(1.0005)^{2048}<3,$ this calculation should be well within you students' abilities.

Answer 4

It is the same as $2018>(1+\frac{1}{2018})^{2018}$, so we only need to show that $$a_n=\left(1+\frac{1}{n}\right)^n$$ is bounded. There are many proofs of this. Here is one possibility.

Let $b_n=(1+\frac{1}{n})^{n+1}$. We observe that $b_n$ is decreasing since $$\begin{aligned}b_{n-1}=\left(\frac n{n-1}\right)^n&=\left[\text{Geometric mean of }\underbrace{\frac n{n-1},\ldots,\frac{n}{n-1}}_n,1\right]^{n+1}\\ &\ge \left[\text{Harmonic mean of }\underbrace{\frac n{n-1},\ldots,\frac{n}{n-1}}_n,1\right]^{n+1}\\ &=\left(\frac{n+1}n\right)^{n+1}=b_n.\end{aligned}$$ Hence, $a_n\le b_n\le b_1=4$.

However, you may need to make them believe that geometric mean is larger than harmonic mean.

Answer 5

We wish to show that

$$\left(2018\right)^{2019}>\left(2019\right)^{2018}$$

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METHODOLOGY $1$:

This is equivalent to showing

$$2018>\left(1+\frac1{2018}\right)^{2018}\tag1$$

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Let us generalize $(1)$ and analyze the sequence $\left(1+\frac1n\right)^n$.

We first note that for $n\ge 2$

$$\left(1+\frac1n\right)^n<\frac1{\left(1-\frac1n\right)^n}\tag2$$

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Furthermore, since the sequence $\left(1-\frac1n\right)^n$ is monotonically increasing, which we show using only Bernoulli's Inequality in the Appendix to this solution, we have

$$\left(1-\frac1n\right)^n\ge \left(1-\frac12\right)^2=\frac14\tag 3$$

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Using $(3)$ in $(2)$ reveals that

$$\left(1+\frac1n\right)^n<4$$

whence we obtain the result $n>\left(1+\frac1n\right)^n$ for $n>4$. Setting $n=2018$ yields the inequality in $(1)$.

And we are done.

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METHODOLOGY $2$

This is equivalent to showing

$$\log(2018)>2018\log\left(1+\frac1{2018}\right)$$

From THIS ANSWER, clearly we have $2018\log\left(1+\frac1{2018}\right)<1$. In addition, it is trivial to see that $\log(2018)>1$.

And we are done!

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APPENDIX: Proof that $\left(1-\frac1n\right)^n$ monotonically decreases

Let $a_n=\left(1-\frac1n\right)^n$. Then, for $n\ge 2$, the ratio $\frac{a_{n+1}}{a_n}$ is given by

$$\begin{align} \frac{a_{n+1}}{a_n}&=\frac{\left(1-\frac1{n+1}\right)^{n+1}}{\left(1-\frac1n\right)^n}\\\\ &=\left(1-\frac1n\right)\left(\frac{1-\frac1{n+1}}{1-\frac1n}\right)^{n+1}\\\\ &=\left(1-\frac1n\right)\left(1+\frac{1}{n^2-1}\right)^{n+1}\tag {A1}\\\\ &\ge \left(1-\frac1n\right)\left(1+\frac{n+1}{n^2-1}\right)\tag{A2}\\\\ &=1 \end{align}$$

Hence, we see that $a_{n+1}\ge a_n$ for $n\ge 2$. And clearly $a_2\ge a_1=0$ so that $a_n$ is monotonically increasing for $n\ge 1$, which completes the proof.

Note in going from $(A1)$ to $(A2)$ we used Bernoulli's Inequality.

Answer 6

Seen that $\,x\mapsto x\,$ grows considerably faster than $\,x\mapsto\ln x\,$ and
when accepting more or less (without calculus at any rate) that $\,x\,/\ln x\,$ is strictly increasing, then $$\frac{2018}{\ln 2018} \:<\: \frac{2019}{\ln 2019}\,.$$

Answer 7

Borrowing from a recent answer of mine (it proves more than is needed here, and it avoids using the Binomial Theorem, but it does use induction, and it is arguably a bit too complicated, even for its original purpose, so I've simplified it considerably for the present application):

Define $$a_n = \left(1+\frac{1}{n}\right)^n \quad (n \geqslant 1). $$

If $x \geqslant y > 0$, and $n$ is a positive integer, then $$ x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1}) \geqslant n(x - y)y^{n-1}. $$ This can be convincingly justified without an explicit use of induction, just a plausible use of ellipsis.

Therefore, for $n > 1$, \begin{align*} a_n - a_{n-1} & = \left(1+\frac{1}{n}\right)^n \! - \left(1+\frac{1}{n-1}\right)^{n-1} \\ & = \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \left[ \left(1+\frac{1}{n-1}\right)^{n-1} \!\! - \left(1+\frac{1}{n}\right)^{n-1}\right] \\ & \leqslant \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{1}{n^2}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{a_n}{(n+1)^2}, \end{align*} whence $$ a_n \leqslant a_{n-1}\left(1 - \frac{1}{(n+1)^2}\right)^{-1} \quad (n > 1). $$ But, for $n > 1$, $$ \left(1 - \frac{1}{(n+1)^2}\right)^{-1} \!\! = 1 + \frac{1}{(n+1)^2 - 1} < 1 + \frac{1}{n-1} = \frac{n}{n-1}, $$ therefore $$ \frac{a_n}{n} < \frac{a_{n-1}}{n-1} \quad (n > 1), $$ and - again without explicit use of induction - one can deduce: $$ \frac{a_n}{n} \leqslant \frac{a_3}{3} = \frac{64}{81} < 1 \quad (n \geqslant 3), $$ whence: $$ (n + 1)^n < n^{n+1} \quad (n \geqslant 3). $$

Answer 8

gt6989b's answer has shown that it suffices to show that $(1 + 1/n)^n < n$ when $n = 2018$. In fact this is true for $n \ge 3$. If you're allowing the "usual" binomial theorem (with positive integer exponents), then you can show this as follows:

$$ \left( {1 + {1 \over n}} \right)^n = \sum_{k=0}^n {n \choose k} \left( {1 \over n}\right)^k = 2 + \sum_{k=2}^n {{n \choose k} \over n^k}. $$

Now I claim that, for $k \ge 3$, $${{n \choose k} \over n^k} < \left( {1 \over 2} \right) ^{k-1}.$$

First observe that $${n \choose k} = {n(n-1) \cdots (n-k+1) \over k!} < {n^k \over k!} $$ and so $$ {{n \choose k} \over n^k} < {1 \over k!} $$ But the denominator $k!$ has a factor of 2 and $k-2$ factors which are greater than 2, so that denominator is at least $2^{k-1}$.

Applying this inequality to the first displayed equation you get $$ \left( {1 + {1 \over n}} \right)^n = 2 + \sum_{k=2}^n {1 \over 2^{k-1}} $$ and summing the geometric series gives $(1 + 1/n)^n < 3$.

Answer 9

Notice that $2018^x > x+1$ for any $x>0$ (plot each function if need be to convince your high school students).

Then $x = \frac{2019}{2018} -1 > 0$ gives

$$\frac{2018^{\frac{2019}{2018}}}{2018} > \frac{2019}{2018}$$

which gives the desired result after clearing denominators and taking each side to the $2018$ power.

Note: This argument is more commonly used on comparing $e^2$ to $2^e$.

Answer 10

Partial answer: we can show using combinatorics that $(n+1)^n < n^n \cdot (n+1)$ for $n > 1$.

If you have a string of $n$ beads that can have $n+1$ colors $c_1, \ldots c_{n+1}$, there must be at least $1$ color that does not appear in the string. Shifting the colors to fill in the gap, we obtain a string of length $n$ where each bead is colored $c_1$ to $c_n$. The string we obtain is not unique; there are $n+1$ ways we could have shifted the colors to obtain that string. Hence the weak inequality. The strict inequality holds because the LHS is not divisible by $n$.

I hope this will inspire someone to prove that $(n+1)^n < n^n \cdot n$.

Answer 11

Note that $2019^{2048}<2018^{2049}$ implies that $$2019^{2018}2019^{30}=2019^{2048}<2018^{2049}<2018^{2019}2019^{30},$$ which implies that $2019^{2018}<2018^{2019}$. We look at $2048$ in the exponent because it is a power of $2$.

Claim: $2019^{2048}<2018^{2049}$.

Proof: $$2019^{2048}-2018^{2048}=$$ $$(2019^{1024}+2018^{1024})(2019^{512}+2018^{512})...(2019^{2}+2018^{2})(2019+2018)(2019-2018)$$ by repeatedly factoring differences of squares. Each term of the form $2019^i+2018^i<2 \cdot 2019^i$, so taking each of these inequalities into account, we get that $$2019^{2048}-2018^{2048}<2^{10}\cdot 2019^{2047}$$ since $1+2+4+8+...+512+1024=2^{11}-1$. Then we combine terms with like bases to get that $$2019^{2048}-1024\cdot 2019^{2047}=995\cdot 2019^{2047}<2018^{2048}.$$ We now multiply both sides by $2018$ to get $$2019^{2048}<995\cdot 2018\cdot 2019^{2047}<2018^{2049},$$ which is the desired result.

Answer 12

Late answer but I think worth noting:

A possible way is using the inequality between the geometric and harmonic mean:

• $\sqrt[n]{a_1 \cdots a_n} \geq \frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}}$ for $a_1, \ldots , a_n > 0$.

Now, we use $$2018^{2019}> 2019^{2018} \Leftrightarrow \sqrt[2018]{2018} > 1+\frac{1}{2018}$$

and show the inequality on the RHS:

\begin{eqnarray*} \sqrt[2018]{2018} & = & \sqrt[2018]{2 \cdot 1009 \cdot 1^{2016}} \\ & \stackrel{GM-HM}{\geq} & \frac{2018}{\frac{1}{2} + \frac{1}{1009} + 2016}\\ & = & 1+ \frac{2-\frac{1}{2} - \frac{1}{1009}}{\frac{1}{2} + \frac{1}{1009} + 2016} \\ & > & 1+ \frac{1}{2018} \\\end{eqnarray*}

Answer 13

Hint:

From

$$n^{1/n}>(n+1)^{1/(n+1)}$$

we draw

$$n^{1/n-1/(n+1)}=\left(\frac{n+1}n\right)^{1/(n+1)},$$

$$n>\left(1+\frac1n\right)^n.$$

If we can show that $\left(1+\dfrac1n\right)^n$ is bounded above, say by $3$, we are done.

Answer 14

Rewrite $$\begin{array} {rll} 2018 \cdot (2018)^{2018} &> 2019^{2018}\\ 2018 &\gt (1+1/2018)^{2018} \\ \ln(2018) &\gt 2018 \cdot \ln(1+1/2018) \\ & \qquad= 2018\cdot(1/2018-1/2018^2/2 + ...)\\ & \qquad \qquad \text{by Mercatorseries expansion of the } \ln() \end{array}$$ leading to $$ \ln(2018) \gt 1-1/2018/2 + 1/2018^2/3 - ... + ... $$ which is obviously true.

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However, I don't know whether we can assume a highschooler would know the Mercator-series for the natural logarithm?

Answer 15

This answer is a fusion of answers given by "@gt6989b" and "@saulspatz". The answers from "@user328442" is intuitively plausible. Thank you all for your responses that I found pertinent with "+1". I would especially like to thank the three of you mentioned above.

I'll write the answer in a friendly way for a 'High school' student. So it's a bit pedantic and long. I insist on putting this answer because there are subtleties in the numerical approximation. Subtleties that are circumvented by the "@saulspatz" user approach.

The strategy of ​​the proof is:

• If we proof $\quad\dfrac{2019^{2018}}{{2018}^{2019}}>1\quad$ then $\quad 2019^{2018} > {2018}^{2019}$.

• If we proof $\quad\dfrac{2019^{2018}}{{2018}^{2019}}<1\quad$ then $\quad 2019^{2018} < {2018}^{2019}$.

Therefore, we must estimate the fraction $$\dfrac{2019^{2018}}{{2018}^{2019}}$$

Note that $$ \frac{2019^{2018}}{{2018}^{2019}} = \frac{2019^{2018}}{2018\cdot {2018}^{2018}} = \frac{1}{2018}\cdot \frac{2019^{2018}}{ {2018}^{2018}} = \frac{1}{2018}\cdot \left(\frac{2019}{2018}\right)^{2018} \\ \quad \\ = \frac{1}{2018}\cdot \left(\frac{2018+1}{2018}\right)^{2018} = \frac{1}{2018}\cdot \left(\frac{2018}{2018}+\frac{1}{2018}\right)^{2018} = \frac{1}{2018}\cdot \left(1+\frac{1}{2018}\right)^{2018} $$ We will prove that $$ \frac{1}{2018}\cdot \left(1+\frac{1}{2018}\right)^{2018}<1 $$ In fact, \begin{align*} \frac{1}{2018}\cdot \left(1+\frac{1}{2018}\right)^{2018} \leq & \frac{1}{2018}\cdot \left(1+\frac{1}{2000}\right)^{2018} \\ \leq & \frac{1}{2018}\cdot \left(1+\frac{1}{2000}\right)^{2048} \\ = & \frac{1}{2018}\cdot (1,0005)^{2^{11}} \\ = & \frac{1}{2018}\cdot (1.00100025)^{2^{10}} \\ \leq & \frac{1}{2018}\cdot (1.0015)^{2^{10}} \\ = & \frac{1}{2018}\cdot (1.00300225)^{2^{9}} \\ \leq & \frac{1}{2018}\cdot (1.0035)^{2^{9}} \\ = & \frac{1}{2018}\cdot (1.00701225)^{2^{8}} \\ \leq & \frac{1}{2018}\cdot (1.0075)^{2^{8}} \\ = & \frac{1}{2018}\cdot (1.01505625)^{2^{7}} \\ \leq & \frac{1}{2018}\cdot (1.0155)^{2^{7}} \\ = & \frac{1}{2018}\cdot (1.03124025)^{2^{6}} \\ \leq & \frac{1}{2018}\cdot (1.035)^{2^{6}} \\ = & \frac{1}{2018}\cdot (1.071225)^{2^{5}} \\ \leq & \frac{1}{2018}\cdot (1.075)^{2^{5}} \\ = & \frac{1}{2018}\cdot (1.155625)^{2^4} \\ \leq & \frac{1}{2018}\cdot (1.175)^{2^4} \\ = & \frac{1}{2018}\cdot (1.380625)^{2^3} \\ \leq & \frac{1}{2018}\cdot 2^{2^3} \\ =& \frac{1}{2018}\cdot 256 <1 \end{align*}