The limit of (1+1/n)^n as n->infinity

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Say you have borrowed on coin, of whatever domination, from a loan shark that takes 100% interest. After one year you will pay back 1+1=2 coins.

Now let us say the loan shark uses compounded interest to be compounded every month. That means that you have to pay 2.6 coins:

$\displaystyle {{\left( {1+\frac{1}{{12}}} \right)}^{{12}}}\approx 2.6$

(Ok. This is a very stupid way to do the calculation, but it is how it is done. The question is why one don’t calculate the monthly interest rate as the 12^th root of 2…. but anyhow…)

Say we instead compound daily. This would give us

$\displaystyle {{\left( {1+\frac{1}{{365}}} \right)}^{{365}}}\approx 2.71$

It looks like we are approaching some limit. So let us examine the limit

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}$

We first expand this using the binominal expansion

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {1+\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{{\left( {\frac{1}{n}} \right)}}^{1}}+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{{\left( {\frac{1}{n}} \right)}}^{2}}+\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right){{{\left( {\frac{1}{n}} \right)}}^{3}}+...} \right)$

This gives us

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {1+n\cdot \frac{1}{n}+\frac{{n\left( {n-1} \right)}}{{2\cdot 1}}\frac{1}{{{{n}^{2}}}}+\frac{{n\left( {n-1} \right)\left( {n-2} \right)}}{{3\cdot 2\cdot 1}}\frac{1}{{{{n}^{3}}}}+...} \right)$

Some cancelling and expansion later we have

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {1+1+\left( {\frac{1}{{2!}}-\frac{1}{{2!n}}} \right)+\left( {\frac{1}{{3!}}-\frac{{...n}}{{2!{{n}^{3}}}}+\frac{{...}}{{2!{{n}^{2}}}}} \right)+...} \right)$

Here the … in the numerators stand for some constants we don’t care about. This because all but the first term in each bracket will approach 0 as n goes toward infinity. We can do the same for the terms after the last …

This gives us

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}=1+\frac{1}{{1!}}+\frac{1}{{2!}}+\frac{1}{{3!}}+\frac{1}{{4!}}+...$

This will obviously be larger than 2 since we add positive numbers to the two first terms, 1+1.

Secondly, we have that 2!=2·1<2², and 2!=3·2·1<2², and so on, and we thus have that

$\displaystyle 2<1+\frac{1}{{1!}}+\frac{1}{{2!}}+\frac{1}{{3!}}+\frac{1}{{4!}}+...<1+\frac{1}{{{{2}^{0}}}}+\frac{1}{{{{2}^{1}}}}+\frac{1}{{{{2}^{2}}}}+\frac{1}{{{{2}^{3}}}}+...$

And we know that

$\displaystyle \frac{1}{{{{2}^{0}}}}+\frac{1}{{{{2}^{1}}}}+\frac{1}{{{{2}^{2}}}}+\frac{1}{{{{2}^{3}}}}+...=2$

And thus that

$\displaystyle 2<1+\frac{1}{{1!}}+\frac{1}{{2!}}+\frac{1}{{3!}}+\frac{1}{{4!}}+...<3$

I.e., the series converges to a value e, called Euler’s number. So, we have that

$e=\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{1}{n}} \right)}^{n}}=1+\frac{1}{{1!}}+\frac{1}{{2!}}+\frac{1}{{3!}}+\frac{1}{{4!}}+...$

You can find a calculator calculating this with quite many more decimals here.

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Last modified: Nov 2, 2023 @ 19:39