2 vs. 1.005^200

Question

Without using a calculator or a computer can you determine which of these two numbers is bigger: $2$ or $1.005^{200}$ ?

I saw this puzzle in a YouTube video, which I will post later.

Answer 1

We can write, $$\begin{align}1.005^{200}&=\left(1+\frac1{200}\right)^{200}\\&=\underbrace{\left(1+\frac1{200}\right)\left(1+\frac1{200}\right)\left(1+\frac1{200}\right)\cdots\left(1+\frac1{200}\right)}_{200\text{ terms}}\end{align}$$ And $$\begin{align}2&=\frac{201}{200}\frac{202}{201}\frac{203}{202}\cdots\frac{400}{399}\\&=\underbrace{\left(1+\frac1{200}\right)\left(1+\frac1{201}\right)\left(1+\frac1{202}\right)\cdots\left(1+\frac1{399}\right)}_{200\text{ terms}}\end{align}$$ As the denominator gets bigger, the number gets smaller. So we can clearly see $$\boldsymbol{1.005^{200}\gt2}$$

Answer 2

This is quite easy if you know

the binomial theorem

or

compound interest.

The largest of the two numbers is

$1.005^{200}$. By the binomial theorem $$(a+b)^{200} = a^{200} + \binom{200}{1} a^{199}b^1 + \binom{200}{2} a^{198}b^2 + ...$$ Since $1.005= 1+\frac{1}{200}$, we can put $a=1$ and $b=\frac{1}{200}$ to find that $$\left(1+\frac{1}{200}\right)^{200} = 1^{200} + 200\cdot 1^{199}\left(\frac{1}{200}\right)^1 + ... = 2 + ...$$

Another way to see it, is by thinking of compound interest. Multiplying by $1.005$ is a half percent increase, and $200$ times such an increase is a $100\%$ increase if you do not take compound interest into account. So it doubles without the compound interest, and more than doubles with the compound interest. So multiplying by $1.005^{200}$ is more than multiplying by $2$.

Answer 3

Multiplying a number $x$ by $1.005$ is equivalent to adding $0.005x$ to $x$. When $x > 1$, $0.005x > 0.005$. Therefore, when $1.005$ is multiplied by itself, the result is greater than $1.005 + 0.005$, when it's multiplied by itself twice, the result is greater than $1.005 + 0.005 + 0.005$, and so on. $1.005^{200}$, or $1.005$ multiplied by itself $199$ times, is therefore greater than $0.005$ added to $1.005$ 199 times. In other words, $1.005^{200} > 1.005 + 0.005 \cdot 199 = 1.005 + 0.995 = 2$.

Something I noticed when I double-checked my answer:

$1.005^{200}$ is quite close to Euler's constant since, when $n$ tends to infinity, $\left(1 + \frac{1}{n}\right)^n$ approaches $e$, and $1.005^{200}$ equals $\left(1 + \frac{1}{n}\right)^n$ where $n = 200$.

Based on the observation above and a comment by Dmitry Kamenetsky, here's another solution to the puzzle:

When $n > 0$, $\left(1 + \frac{1}{n}\right)^n$ is strictly increasing. Below is my incomplete attempt to prove that, but I'm not great at maths so please tell me if there's anything I should correct.

$$\begin{align*} f(n) &= \left(1 + \frac{1}{n}\right)^n \\ f'(n) &= \left(1 + \frac{1}{n}\right)^n \cdot \left(\ln\left(1 + \frac{1}{n}\right) n\right)' \\ &= \left(1 + \frac{1}{n}\right)^n \cdot \left(\left(\ln\left(1 + \frac{1}{n}\right)\right)' \cdot n + (n)' \cdot \ln(1 + \frac{1}{n})\right) \\ &= \left(1 + \frac{1}{n}\right)^n \cdot \left( \frac{1}{1 + \frac{1}{n}} \cdot (1 + \frac{1}{n})' \cdot n + 1 \cdot \ln(1 + \frac{1}{n})\right) \\ &= \left(1 + \frac{1}{n}\right)^n \cdot \left( \frac{n}{n + 1} \cdot (n^{-1})' \cdot n + \ln\left(1 + \frac{1}{n}\right)\right) \\ &= \left(1 + \frac{1}{n}\right)^n \cdot \left( \frac{n}{n + 1} \cdot \left(-\frac{1}{n^2}\right) \cdot n + \ln\left(1 + \frac{1}{n}\right)\right) \end{align*}$$

If both factors in $f'(n)$ are greater than $0$ when $n > 0$, $f'(n)$ is greater than $0$ when $n > 0$ and $f(n)$ is strictly increasing when $n > 0$.

Proving that the first factor, $\left(1 + \frac{1}{n}\right)^n$, is greater than $0$ is quite simple. $\frac{1}{n}$, and therefore also $1 + \frac{1}{n}$, are greater than $0$ when $n > 0$, and raising a number greater than $0$ to a real power yields a number greater than $0$.

I was not able to prove that the second factor is always greater than $0$ for $n > 0$, but @isaacg did in his own answer, so please check out that one.

Anyway, assuming $f(n)$ is strictly increasing when $n > 0$, $f(200)$, which equals $1.005^{200}$, is greater than $f(1)$, which equals $2$.

Answer 4

We can also use AM-GM, the inequality between the arithmetic and geometric means, applied to 199 1s and one 2:

$1.005 = \frac{\overbrace{1+1+\cdots+1+1}^{199 \text{ terms}}+2}{200}>\sqrt[200]{\underbrace{1\times 1\times\cdots\times 1\times 1}_{199 \text{ terms}}\times 2} = \sqrt[200] 2$

Answer 5

There is binominal formula, first two terms of which is impossible to forget:

$$\displaystyle (1+x)^{n}=1+nx+...$$

so clearly

$$1.005^{200} > 1 + 1$$

Answer 6

For a quick estimation of doubling, we can apply

the rule of 72

The "initial capital" is 1, and the "interest rate is 0.005, or 0.5%.

$$\frac{72}{0.5} = 72 \times 2 = 144$$

Therefore

$$1.005^{144} ≈ 2 < 1.005^{200}$$

Answer 7

I've got the simplest answer.

$(1+x)^2=1^2+2x+x^2 > 1+2x$ if $x>0$, and this obviously extends to $(1+x)^n > 1+nx$.

So $(1+0.005)^{200}>1+200*0.005=1+1=2$.

Answer 8

I feel like a lot of these answers are incredibly complicated... All you need is one simple multiplication.

1.005^200 = 1 + 0.005 * 200 + that compounding stuff, let's call it C. 0.005 * 200=1. So 1.005^200 = 1+1+C. C "obviously" (if you know how compounding works that is) is >0, so 1+1+C > 2.

Answer 9

I wanted to continue @Peter's approach of showing that $f(n) = (1+1/n)^n$ is increasing over the interval $n>0$. Peter showed that

$$f'(n) = (1 + \frac{1}{n})^n \cdot ( \frac{n}{n + 1} \cdot (-\frac{1}{n^2}) \cdot n + \ln(1 + \frac{1}{n})) $$ Because $(1 + \frac{1}{n})^n$ is clearly positive, we just need to show that $$( \frac{n}{n + 1} \cdot (-\frac{1}{n^2}) \cdot n + \ln(1 + \frac{1}{n})) > 0$$

Let's simplify this term:

$$ \frac{n}{n + 1} \cdot (-\frac{1}{n^2}) \cdot n + \ln(1 + \frac{1}{n}) = -\frac{1}{n+1} + \ln(1 + \frac{1}{n})$$

So all we need to show is that

$$\ln(1+\frac{1}{n}) \ge \frac{1}{n+1}$$

To prove this, let us apply

the lower bound of the natural logarithm, which states that $$ \ln x \ge 1-\frac{1}{x}$$ Applying this with $x=1+\frac{1}{n}$, we find that $$ \ln (1+\frac{1}{n}) \ge 1 - \frac{1}{1+\frac{1}{n}} = 1 - \frac{n}{n+1} = \frac{1}{n+1}$$

This completes the proof that $f'(n) > 0$, as desired.

Answer 10

As others have mentioned, a simple solution is to look at the first few terms of the binomial expansion of $\left(1+ \frac{1}{200}\right)^{200}$: $$1 + \frac{200}{1}\cdot\frac{1}{200} + \frac{200\cdot199}{2}\cdot \frac{1}{200^2} + \cdots$$ which is clearly $>2$.

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It has also been mentioned that for large $x$, $(1+ \frac{1}{x})^{x} \approx e$. In fact, it can be shown that for $x>0$, $$\left(1+ \frac{1}{x}\right)^{x} < e < \left(1+ \frac{1}{x}\right)^{x+1}$$ and hence $$e\approx\left(1+ \frac{1}{x}\right)^{x+1/2}$$ We can use this approximation to give a better estimate of $a=\left(1+ \frac{1}{200}\right)^{200}$, without a calculator.

We have $$e\approx\left(1+ \frac{1}{200}\right)^{200+1/2}$$ So $$a\approx e \left(1+ \frac{1}{200}\right)^{-1/2}$$

For small $u$, $\frac{1}{1+u}\approx 1-u$.
Also, $(1+u)^2\approx 1+2u$, so $\sqrt{1+u}\approx 1+u/2$.
Thus, $\left(1+ \frac{1}{200}\right)^{-1/2} \approx \left(1- \frac{1}{400}\right)$ and so $a\approx e\left(1- \frac{1}{400}\right)$

Now, we just need an approximation of $e$. Any student of calculus should know that $e\approx2.71828$, but it's easy enough to calculate a few decimals by hand from the Taylor series: $$e^x=\sum_{i=0}^{\infty} \frac{x^i}{i!}$$ The first few sums for $e$ are

i	S(i)
0	1/1
1	2/1
2	5/2
3	16/6
4	65/24
5	326/120
6	1957/720
7	13700/5040

The last row gives us $e\approx 2.71825$. Subtracting $2.71825/400$ gives $a\approx 2.71145$. A high precision calculation of $a$ gives $2.7115171229293747985490$, so that approximation isn't too shabby.

FWIW, apart from the final high precision calculation, I performed all of the arithmetic by hand, but I checked it with a calculator.

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Incidentally, there's an alternative way to calculate $e$ related to $e\approx\left(1+ \frac{1}{x}\right)^{x+1/2}$. It's more convenient to make the power an integer: $e\approx\left(1+ \frac{1}{x-1/2}\right)^x$, and if $x$ is a binary power we can do exponentiation by repeated squaring.

We can do even better. Let $$e=\left(1+ \frac{1}{x-1/2 + y}\right)^x$$ Then we can find $y$ from the continued fraction expansion of $e^{1/x}$.
We get $y=$ $$\cfrac{1}{12x + \cfrac{1}{5x + \cfrac{1}{28x + \cfrac{1}{9x + \cdots\vphantom{\cfrac{1}{1}}}}}}$$ See https://oeis.org/A110185 for more coefficients, which come from an interleaved pair of arithmetic progressions. Using just the terms shown above for $y$ with $x=256$, we get $e=2.7182818284590452353602874713526627$, which is accurate to 110 bits.

Answer 11

A solution using the Mean Value Theorem:

We wish to determine whether $1.005^{200} - 2$ is positive or negative, or equivalently, whether $1005^{200} - 2 \times 1000^{200}$ is positive or negative. We may isolate this quantity as $(1005^{200} - 1000^{200}) - 1000^{200}$, and the part within the parentheses is of the form $f(1005) - f(1000)$ for $f(x) = x^{200}$, which is what gave me the idea.

The mean value theorem yields $1005^{200} - 1000^{200} = f(1005) - f(1000) = 5 \times f'(\xi)$ for some $1000 < \xi < 1005$, and thus $$ 1005^{200} - 1000^{200} = 5 \times 200 \times \xi^{199} = 1000 \times \xi^{199}. $$

The fact that $5 \times 200 = 1000$ is fortunate, and allows us to see that $1005^{200} - 1000^{200} = 1000 \times \xi^{199} > 1000^{200}$, and thus $(1005^{200} - 1000^{200}) - 1000^{200} > 0$, whence $1.005^{200} - 2 > 0$ and hence $1.005^{200} > 2$.

Answer 12

Since (1+a)²=1+2a+a²
As a²>0,
(1+a)²>1+2a
1.005²>1.01
1.005⁴>1.01²>1.02

Similarly,

(1+a)⁵⁰>1+50a
1.02⁵⁰>1+50(0.02)=1+1=2

Combining the results:

1.005²⁰⁰=(1.005⁴)²⁰>2

Answer 13

Another proof of Bernoulli's inequality:

For $n > 1$, $f(x) = (1 + x)^n$ is strictly convex over $x > -1$ as its second derivative is $(n-1)(n)(x+1)^{n-2} > 0$.

So by convexity, $f$ lies above its tangent line at $x=0$, that is for $x \ne 0$, $(1+x)^n = f(x) > f(0) + f'(0)x = 1 + nx$. The puzzle follows from $x = 1/200, n=200$.