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

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Say you have borrowed one 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 doesn’t calculate the monthly interest rate as the 12th 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 is 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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