The Euler Number $e$

Now let's consider sequence ${x}_{{n}}={{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}$ and try to find its limit.

It is not very clear whether this sequence is monotonic or not.

So, to make sure that this sequence is increasing let's rewrite sequence using binom of Newton with ${a}={1}$ and ${b}=\frac{{1}}{{n}}$:

${{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}={1}+{{C}_{{n}}^{{1}}}\frac{{1}}{{n}}+{{C}_{{n}}^{{2}}}\frac{{1}}{{{n}}^{{2}}}+{{C}_{{n}}^{{3}}}\frac{{1}}{{{n}}^{{3}}}+\ldots+{{C}_{{n}}^{{k}}}\frac{{1}}{{{n}}^{{k}}}+\ldots+{{C}_{{n}}^{{{n}-{1}}}}\frac{{1}}{{{n}}^{{{n}-{1}}}}+\frac{{1}}{{{n}}^{{n}}}$

Or ${{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}={1}+\frac{{n}}{{n}}+\frac{{{n}{\left({n}-{1}\right)}}}{{2}}\frac{{1}}{{{n}}^{{2}}}+\frac{{{n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}}}{{6}}\frac{{1}}{{{n}}^{{3}}}+\ldots+$

$+\frac{{{n}{\left({n}-{1}\right)}\ldots{\left({n}-{k}+{1}\right)}}}{{{k}!}}\frac{{1}}{{{n}}^{{k}}}+\ldots+{n}\frac{{1}}{{{n}}^{{{n}-{1}}}}+\frac{{1}}{{{n}}^{{n}}}={1}+{1}+\frac{{1}}{{2}}{\left({1}-\frac{{1}}{{n}}\right)}+\frac{{1}}{{6}}{\left({1}-\frac{{1}}{{n}}\right)}{\left({1}-\frac{{2}}{{n}}\right)}+\ldots+$

$+\frac{{1}}{{{k}!}}{\left({1}-\frac{{1}}{{n}}\right)}{\left({1}-\frac{{2}}{{n}}\right)}\ldots{\left({1}-\frac{{{k}-{1}}}{{n}}\right)}+\ldots+\frac{{1}}{{{n}!}}{\left({1}-\frac{{1}}{{n}}\right)}{\left({1}-\frac{{2}}{{n}}\right)}\ldots{\left({1}-\frac{{{n}-{1}}}{{n}}\right)}$.

Now, if we consider expansion of ${x}_{{{n}+{1}}}$ i.e. increase ${n}$ by 1 then one new ${\left({n}+{2}\right)}$ -th (positive) member will be added and each of written ${n}+{1}$ members will increase, because every factor of the form ${1}-\frac{{s}}{{n}}$ will be substituted with greater factor ${1}-\frac{{s}}{{{n}+{1}}}$.

Therefore, ${x}_{{{n}+{1}}}>{x}_{{n}}$ and sequence is increasing.

Now, let's show that this sequence is bounded from above.

Since every factor inside parenthesis in the expansion of ${{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}$ is less than 1, i.e. every factor of the form ${1}-\frac{{s}}{{n}}<{1}$ then ${x}_{{n}}={{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}<{2}+\frac{{1}}{{{2}!}}+\frac{{1}}{{{3}!}}+\ldots+\frac{{1}}{{{n}!}}$.

Next since ${n}!={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}>{2}\cdot{2}\cdot{2}\ldots\cdot{2}={{2}}^{{{n}-{1}}}$ then $\frac{{1}}{{{n}!}}<\frac{{1}}{{{{2}}^{{{n}-{1}}}}}$.

This means that ${x}_{{n}}<{2}+\frac{{1}}{{{2}!}}+\frac{{1}}{{{3}!}}+\ldots+\frac{{1}}{{{n}!}}<{2}+\frac{{1}}{{2}}+\frac{{1}}{{{{2}}^{{2}}}}+\ldots+\frac{{1}}{{{{2}}^{{{n}-{1}}}}}$.

${S}=\frac{{1}}{{2}}+\frac{{1}}{{{2}}^{{2}}}+\ldots+\frac{{1}}{{{2}}^{{{n}-{1}}}}$ is geometric sequence with the first member ${b}_{{1}}=\frac{{1}}{{2}}$ and common ratio ${q}=\frac{{1}}{{2}}$. Note that here we have ${n}-{1}$ summands.

Therefore, ${S}=\frac{{\frac{{1}}{{{2}}^{{n}}}-\frac{{1}}{{2}}}}{{\frac{{1}}{{2}}-{1}}}={1}-\frac{{1}}{{{2}}^{{{n}-{1}}}}<{1}$.

Therefore, ${x}_{{n}}<{2}+{S}<{2}+{1}<{3}$.

Thus, sequence ${x}_{{n}}$ is increasing and bounded from above by 3. According to theorem about monotonic sequence this sequence has finite limit.

This limit is denoted by ${e}$: ${e}=\lim_{{{n}\to\infty}}{{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}$.

It is called Euler's number and is very important for calculus and different applications.

Its decimal representation is ${e}\approx{2.718281828459045}\ldots$.