# 2 vs. 1.005^200

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.

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 $$\boldsymbol{1.005^{200}\gt2}$$

• I've not seen this one before. Very nice! Commented Aug 18, 2023 at 12:08
• Agree. This was very nice. Commented Aug 18, 2023 at 12:12
• This is the nice solution I was looking for. Well done. It is described in this video: youtu.be/CN59vNllv7g Commented Aug 18, 2023 at 12:33
• What an elegant answer! Commented Aug 18, 2023 at 19:44
• It took a moment to realize where the expansion of 2 came from. A helpful step for us less mathematically inclined could be 2=400/200=400/200*201/201*202/202*...*399/399 and then rearranging. Commented Aug 21, 2023 at 8:13

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 $$2$$.

• This is clearly the intended solution. It's no coincidence that 0.005 is 1/200. Commented Aug 18, 2023 at 22:10
• Also Bernoulli's inequality, which can be proved without knowing the binomial theorem.
– qwr
Commented Aug 21, 2023 at 19:28

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 $$2$$.

• Nice observation about Euler's constant. Perhaps that alone is enough to find the bigger number. Commented Aug 18, 2023 at 12:35
• @DmitryKamenetsky It is. (Use the right (!) inequality.) Commented Aug 18, 2023 at 13:45
• @DmitryKamenetsky That's probably true! $(1 + frac{1}{n})^n$ is, after all, strictly increasing when $n > 0$, so the result when $n$ equals $200$ must be greater than the result when $n$ equals $1$ (which is $2$). I might add a more formal proof for this to my answer later :) Commented Aug 18, 2023 at 14:15
• Nice approach! I completed your proof that (1+1/n)^n is strictly increasing, see my answer. Commented Aug 18, 2023 at 20:11
• Thank you, @isaacg! I'll edit my answer to mention yours :) Commented Aug 19, 2023 at 6:01

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

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

• that is very neat! Commented Aug 18, 2023 at 23:32

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

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

so clearly

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

• This is true as long as all other terms are positive, which is true in this specific case. Commented Aug 20, 2023 at 16:26

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

$$1.005^{144} ≈ 2 < 1.005^{200}$$

• I am impressed that you were able to find a close approximation to 2 with rule of 72. Although rule of 70 gives even a better approximation. Commented Aug 18, 2023 at 12:40

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

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

• en.wikipedia.org/wiki/Bernoulli%27s_inequality Commented Aug 18, 2023 at 20:41
• This is essentially the first half of @Jaap's answer. Commented Aug 18, 2023 at 21:21
• Ok you did in one step that which took me two. Commented Aug 19, 2023 at 17:47
• When you say "this obviously extends to...", do you mean "I could prove by induction that..."?
– Stef
Commented Aug 20, 2023 at 13:37
• @Stef One can use binomial theorem again to prove that... Commented Aug 20, 2023 at 14:32

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.

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.

• This also follows from the mean value theorem: $\log 1+ \frac 1 n = \log n+1 - \log n = \frac 1 \nu$ for some $\nu$ strictly between $n$ and $n+1$. Commented Aug 19, 2023 at 0:40

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 $$>2$$.

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.

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.

• Some fascinating trivia about e. Thank you! Commented Aug 20, 2023 at 11:31
• Glad you like it @Dmitry! You may enjoy math.stackexchange.com/a/1295561/207316 which has Python code to calculate large numbers of digits of e (without using arbitrary precision arithmetic). Commented Aug 20, 2023 at 11:39
• Very cool, thank you. That's the first time I've seen the factorial base used for anything useful. Commented Aug 21, 2023 at 8:37

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 $$1.005^{200} > 2$$.

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

• Hi. I don't quite follow the logic. In the first block I understand that you're showing by induction that $(1 + a)^{(2^n)} > 1 + 2^n a$; but then in the second block you conclude that $(1+a)^{50} > 1 + 50a$, even though 50 is not a power of 2.
– Stef
Commented Aug 20, 2023 at 13:49
• Read some of the other answers. They explain it better and you will get a general feel of the solution. Commented Aug 20, 2023 at 22:28
• I don't see any induction here, just using Bernoulli's inequality twice when it could be used directly on 200.
– qwr
Commented Aug 21, 2023 at 20:53
• I agree. It's not the neatest answer. Commented Sep 2, 2023 at 15:38

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 $$x = 1/200, n=200$$.