Why doesn't substituting very large $n$ into $(1+1/n)^n$ give values approaching Euler's number $e$?

Question

I would like to ask what is the Euler's number $e$? I don't understand it.

What I Know:

• $e$ is $\left(1+\dfrac{1}{n}\right)^n$ as $n$ reaches infinity

• $e$ is $2.718281828\ldots$

Question:

If I input $n = 1\ 000\ 000\ 000\ 000$, I get $2.718523496\ldots$, which is higher than $2.718281828\ldots$.

If I go on and input $n = 1\ 000\ 000\ 000\ 000\ 000$, I get $3.035035207\ldots$ which is way higher than $2.718281828\ldots$.

I think I missed something or I made a mistake.

Did I misunderstand the formula? Is the formula and the $2.718281828$ above only an approximation?

I would really appreciate any explanation, clarification, and corrections. :)

Thank you so much for you time!

Answer 1

Here's an error analysis. If $$a_n=\left(1+\frac1n\right)^n$$ then $$\ln a_n=n\ln\left(1+\frac1n\right)=n\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots\right)=1-\frac1{2n}+\frac1{3n^2}-\cdots.$$ For large $n$, $\ln a_n$ is very close to $$1-\frac1{2n}$$ and so $a_n$ is close to $$e\exp(-1/(2n))=e\left(1-\frac1{2n}+\frac1{8n^2}-\cdots\right).$$ Actually the $1/(8n^2)$ term here is spurious as I neglected the $1/(3n^2)$ term in the expansion of $\ln a_n$. But a crude estimate of $a_n$ is that $$a_n\approx e-\frac{e}{2n}.$$ The error is slightly worse than $1/n$.

Taking $n=10^{12}$ say, you get about $11$ to $12$ correct decimal places. The error you get with the calculator is no doubt due to its lack of precision of its representation of floating point numbers. Probably underflow.

Answer 2

Floating point math in a computer is not the same as real math computation. Back when we used $32$ bit floats, which only gave $23$ bits of mantissa, about $7.2$ decimal digits, it was a problem everybody worried about and large sections of numerical analysis courses concentrated on avoiding the problems of numerical precision. Now that floats are $64$ bits with $53$ bits of mantissa the problem has been greatly reduced, but it still can have a problem. When you raise to a very small power you can think about $(1+\frac 1n)^n=e^{(\log(1+\frac 1n)n)}$ and expand $\log(1+\frac 1n)$ in a Taylor series.