The Definition of Euler's Constant, the Natural Number e
📂Analysis The Definition of Euler's Constant, the Natural Number e Definition The limit of the series below is defined as constant e e e
e : = ∑ n = 0 ∞ 1 n !
e: = \sum \limits_{n=0}^{\infty} \dfrac{1}{n!}
e := n = 0 ∑ ∞ n ! 1
Explanation Even if we do not know what that value is right away, it can be easily shown that the series mentioned above converges to some limit. The partial sum s n s_{n} s n bounded and increasing, converges .
s n = 1 + 1 + 1 2 + 1 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 4 + ⋯ + 1 1 ⋅ 2 ⋅ ⋯ ⋅ n < 1 + 1 + 1 2 + 1 2 ⋅ 2 + 1 2 ⋅ 2 ⋅ 2 + ⋯ 1 2 ⋅ 2 ⋅ ⋯ ⋅ 2 ⏟ n − 1 = 1 + 1 + 1 2 + 1 2 2 + 1 2 3 + ⋯ 1 2 n − 1 < 1 + 1 + 1 < 3
\begin{align*}
s_{n} &= 1 + 1 + \dfrac{1}{2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \cdots + \dfrac{1}{1 \cdot 2 \cdot \cdots \cdot n}
\\ &< 1 + 1 + \dfrac{1}{2} + \dfrac{1}{2 \cdot 2} + \dfrac{1}{2 \cdot 2 \cdot 2} + \cdots \dfrac{1}{\underbrace{2 \cdot 2 \cdot \cdots \cdot 2}_{n-1}}
\\ &= 1 + 1 + \dfrac{1}{2} + \dfrac{1}{2^{2}} + \dfrac{1}{2^{3}} + \cdots \dfrac{1}{2^{n-1}}
\\ &< 1 + 1 + 1 < 3
\end{align*}
s n = 1 + 1 + 2 1 + 2 ⋅ 3 1 + 2 ⋅ 3 ⋅ 4 1 + ⋯ + 1 ⋅ 2 ⋅ ⋯ ⋅ n 1 < 1 + 1 + 2 1 + 2 ⋅ 2 1 + 2 ⋅ 2 ⋅ 2 1 + ⋯ n − 1 2 ⋅ 2 ⋅ ⋯ ⋅ 2 1 = 1 + 1 + 2 1 + 2 2 1 + 2 3 1 + ⋯ 2 n − 1 1 < 1 + 1 + 1 < 3
Theorem lim n → ∞ ( 1 + 1 n ) n = e
\lim _{n \to \infty} \left( 1 + \dfrac{1}{n} \right)^{n} = e
n → ∞ lim ( 1 + n 1 ) n = e
Or
lim n → 0 ( 1 + n ) 1 n = e
\lim _{n \to 0} \left( 1 + n \right)^{\frac{1}{n}} = e
n → 0 lim ( 1 + n ) n 1 = e
It’s fine to define it as such. In the Wade textbook
Proof Let’s assume
s n = ∑ k = 0 n 1 k ! , t n = ( 1 + 1 n ) n
s_{n} = \sum \limits _{k=0}^{n} \dfrac{1}{k!},\quad t_{n}=\left( 1 + \dfrac{1}{n} \right)^{n}
s n = k = 0 ∑ n k ! 1 , t n = ( 1 + n 1 ) n
Binomial Theorem
( x + y ) n = ∑ r = 0 n n C r x n y n − r
(x+y)^{n} = \sum_{r=0}^{n} {_n C _r} x^{n} y^{n-r}
( x + y ) n = r = 0 ∑ n n C r x n y n − r
Then, by the binomial theorem, the following holds.
t n = n C 0 ⋅ 1 + n C 1 1 n + n C 2 1 n 2 + n C 3 1 n 3 + ⋯ + n C n 1 n n = 1 + n 1 n + n C 2 1 n 2 + n C 3 1 n 3 + ⋯ + n C n 1 n n = 1 + 1 + 1 2 ! n ( n − 1 ) 1 n 2 + 1 3 ! n ( n − 1 ) ( n − 2 ) 1 n 3 + ⋯ + 1 n ! n ( n − 1 ) ⋯ 2 ⋅ 1 1 n n = 1 + 1 + 1 2 ! ( 1 − 1 n ) + 1 3 ! ( 1 − 1 n ) ( 1 − 2 n ) + ⋯ + 1 n ! ( 1 − 1 n ) ( 1 − 2 n ) ⋯ ( 1 − n − 1 n )
\begin{align*}
t_{n} &= {_{n}C_{0}} \cdot 1 + {_{n}C_{1}} \dfrac{1}{n}+ {_{n}C_{2}}\dfrac{1}{n^{2}} + {_{n}C_{3}}\dfrac{1}{n^{3}} + \cdots + {_{n}C_{n}}\dfrac{1}{n^{n}}
\\ &= 1 + n \dfrac{1}{n}+ {_{n}C_{2}}\dfrac{1}{n^{2}} + {_{n}C_{3}}\dfrac{1}{n^{3}} + \cdots + {_{n}C_{n}}\dfrac{1}{n^{n}}
\\ &= 1 + 1+ \dfrac{1}{2!}n(n-1)\dfrac{1}{n^{2}} + \dfrac{1}{3!}n(n-1)(n-2)\dfrac{1}{n^{3}} + \cdots + \dfrac{1}{n!}n(n-1)\cdots 2 \cdot 1\dfrac{1}{n^{n}}
\\ &= 1 + 1+ \dfrac{1}{2!}\left(1-\dfrac{1}{n}\right) + \dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \cdots + \dfrac{1}{n!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{n-1}{n}\right)
\end{align*}
t n = n C 0 ⋅ 1 + n C 1 n 1 + n C 2 n 2 1 + n C 3 n 3 1 + ⋯ + n C n n n 1 = 1 + n n 1 + n C 2 n 2 1 + n C 3 n 3 1 + ⋯ + n C n n n 1 = 1 + 1 + 2 ! 1 n ( n − 1 ) n 2 1 + 3 ! 1 n ( n − 1 ) ( n − 2 ) n 3 1 + ⋯ + n ! 1 n ( n − 1 ) ⋯ 2 ⋅ 1 n n 1 = 1 + 1 + 2 ! 1 ( 1 − n 1 ) + 3 ! 1 ( 1 − n 1 ) ( 1 − n 2 ) + ⋯ + n ! 1 ( 1 − n 1 ) ( 1 − n 2 ) ⋯ ( 1 − n n − 1 )
Therefore, ∀ n ∈ N , t n ≤ s n \forall n\in \mathbb{N}, t_{n} \le s_{n} ∀ n ∈ N , t n ≤ s n
lim sup n → ∞ t n ≤ lim sup n → ∞ s n = lim n → ∞ s n = e
\begin{equation}
\limsup \limits_{n \to \infty} t_{n} \le \limsup \limits_{n \to \infty} s_{n} = \lim \limits_{n \to \infty} s_{n} = e
\label{eq1}
\end{equation}
n → ∞ lim sup t n ≤ n → ∞ lim sup s n = n → ∞ lim s n = e
Also, if n ≥ m n \ge m n ≥ m
t n ≥ 1 + 1 + 1 2 ! ( 1 − 1 n ) + 1 3 ! ( 1 − 1 n ) ( 1 − 2 n ) + ⋯ + 1 m ! ( 1 − 1 n ) ( 1 − 2 n ) ⋯ ( 1 − m − 1 n )
t_{n} \ge 1 + 1+ \dfrac{1}{2!}\left(1-\dfrac{1}{n}\right) + \dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \cdots + \dfrac{1}{m!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{m-1}{n}\right)
t n ≥ 1 + 1 + 2 ! 1 ( 1 − n 1 ) + 3 ! 1 ( 1 − n 1 ) ( 1 − n 2 ) + ⋯ + m ! 1 ( 1 − n 1 ) ( 1 − n 2 ) ⋯ ( 1 − n m − 1 )
Note that the RHS is not t m t_{m} t m t n t_{n} t n lim inf n t o ∞ \lim \inf \limits_{n\ to \infty} lim n t o ∞ inf m m m
lim inf n → ∞ t n ≥ 1 + 1 + 1 2 ! + 1 3 ! + ⋯ + 1 m !
\liminf \limits_{n \to \infty} t_{n} \ge 1 + 1+ \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{m!}
n → ∞ lim inf t n ≥ 1 + 1 + 2 ! 1 + 3 ! 1 + ⋯ + m ! 1
Then, taking the limit of m → ∞ m \to \infty m → ∞
lim inf n → ∞ t n ≥ e
\begin{equation}
\liminf \limits_{n \to \infty} t_{n} \ge e
\label{eq2}
\end{equation}
n → ∞ lim inf t n ≥ e
Thus, by ( eq1 ) , ( eq2 ) \eqref{eq1}, \eqref{eq2} ( eq1 ) , ( eq2 )
lim inf n → ∞ t n = e = lim sup n → ∞ t n ⟹ lim n → ∞ t n = e
\liminf \limits_{n \to \infty} t_{n} = e = \limsup \limits_{n \to \infty} t_{n} \implies \lim \limits_{n\to \infty} t_{n} = e
n → ∞ lim inf t n = e = n → ∞ lim sup t n ⟹ n → ∞ lim t n = e
See Also 