Limits

The Taylor series of $e^x$ centered at 0 is

$\sum_{n=0}^{\infty}{\dfrac{x^n}{n!}}$

It is known that this series converges for all x, real or imagenary (This can be proved by the ratio test - try it). The limit test states that if

$\lim_{n\to\infty}{S_n}\neq 0$

Then the series $S_n$ diverges. Therefore,

$\lim_{n\to\infty}{\dfrac{x^n}{n!}}$

Must be $\boxed{0}$ , or else the series would diverge.

Preliminary observations:

• Throughout the proof I'll consider the case $n>a$ because $n\rightarrow \infty$ . WLOG I'll consider $a\in N$
• Proposition: $n!>a!{ (a+1) }^{ n-a }$ ; the proof (*)

Proof:

Proposition $\Longrightarrow \quad 0<\frac { { a }^{ n } }{ n! } <\frac { { a }^{ n } }{ a!{ (a+1) }^{ n-a } }$ $\Longleftrightarrow \quad 0<\lim _{ n\rightarrow \infty }{ \frac { { a }^{ n } }{ n! } } <\lim _{ n\rightarrow \infty }{ \frac { { a }^{ n } }{ a!{ (a+1) }^{ n-a } } }=l$

Let's compute (or calculate?) $l$

$l=\lim _{ n\rightarrow \infty }{ \frac { { a }^{ n } }{ a!{ (a+1) }^{ n-a } } } =\frac { { (a+1) }^{ a } }{ a! } \lim _{ n\rightarrow \infty }{ \frac { { a }^{ n } }{ { (a+1) }^{ n } } =k } \cdot 0=0$

The truth of the thesis follow from the squeeze theorem .

(*) Proof of the preposition: $1\cdot 2\cdot 3\cdot ...\cdot a\cdot (a+1)\cdot ...\cdot n>1\cdot 2\cdot 3\cdot ...\cdot a\cdot { (a+1) }^{ n-a }\quad \Leftrightarrow \quad (a+1)\cdot ...\cdot n>{ (a+1) }^{ n-a }$

$e^{x} = \sum_{k=0}^{\infty} \frac {x^{k}}{k!}$

$U_n < 1$

The limit will have to be $\boxed 0$ .