Write the function $f(x)=e^x$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.

Write the function $$f(x)=e^x$$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.

My solution goes like this:

Taylor's Theorem with Lagrange's form of remainder states that, if $$f$$ is a function defined on an interval $$[a,a+h]$$ where $$h>0$$ and if $$f$$ satisfies the following properties i.e

(i) $$f^{(n-1)}$$ is continuous over $$[a,a+h]$$,

(ii) $$f^{(n-1)}$$ is derivable over $$[a,a+h]$$,

then, we can write, $$f(a+h)=f(a)+hf(a)+\frac {h^2}{2!}f''(a)+\cdots +\frac {h^{n-1}}{(n-1)!}f^{(n-1)}(a)+\frac{h^n}{n!}f^n(a+\theta h),$$ where $$\theta \in (0,1)$$.

If $$h=x$$ and $$a=0$$ we get, the Maclaurin's formula i.e, $$f(x)=f(0)+xf(0)+\frac {x^2}{2!}f''(0)+\cdots +\frac {x^{n-1}}{(n-1)!}f^{(n-1)}(0)+\frac{x^n}{n!}f^n(\theta x),$$ where $$\theta \in (0,1)$$ in the interval $$[0,x].$$

Here, $$f(x)=e^x.$$ Since, $$f$$ everywhere continous and differentiable so, we apply Maclaurin's formula. Also, $$f^n(x)=e^x.$$

We have, $$\lim_{n\to\infty}f(x)=e^x.$$

We note that, $$\lim_{n\to\infty}\frac{x^n}{n!}f^n(\theta x)=\lim_{n\to\infty}\frac{x^n}{n!}e^{\theta x}=0.$$ This is because, $$e^{\theta x}$$ is bounded (for, $$1\leq e^{\theta x}\leq e^x$$) and also, $$\lim_{n\to\infty}\frac{x^n}{n!}=0.$$

We write, $$y_n=f(0)+xf(0)+\frac {x^2}{2!}f''(0)+\cdots +\frac {x^{n-1}}{(n-1)!}f^{(n-1)}(0).$$

So, according to Maclaurin's formula in the interval, $$[0,x]$$ we have, $$f(x)=y_{n-1}+R_n,$$ where $$R_n=\frac{x^n}{n!}f^n(\theta x).$$ Since, $$\{f(x)\}$$ is (a constant sequence and hence,) a convergent sequence, $$\{R_n\}$$ is an convergent sequence as well, so, the sequence (of partial sums) $$y_{n-1}$$ is convergent as well and it's limit exists. This means, $$\lim_{n\to\infty} f(x)=e^x=\lim_{n\to\infty} y_{n-1}+\lim_{n\to\infty}R_n=\lim_{n\to\infty}y_n.$$

So, $$e^x=1+x+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\cdots.$$

As $$x\in\Bbb R$$ is an arbitrary so, the above is the general infinite series representation of $$e^x.$$

However, in a book, this problem is solved in a much complicated way i.e they first consider the sequence, $$\{y_n\}$$ just like I did, and then they used, D'Alambert's Ratio Test, to prove that the series $$1+x+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\cdots$$ is convergent and the rest part was similar to what followed my work i.e taking limits on both sides and all that. But what I want to know, whether my work is still valid or not? This is because I don't think there is any need to invoke any Ratio Test to show that the series is indeed a convergent one as it folows trivially from the convergence of sequence of partial sums $$\{y_n\}.$$ Any help regarding this issue will be highly appreciated.

Here's a picture of the thing given in the book. The book is a regional one.

In particular the root test and ratio tests for convergence of series are based on the convergence of $$1+r+r^2+\dots+r^n+\dots$$ And we can get the partial sums directly via the formula $$1+r+r^2+\dots+r^{n-1}=\frac{1-r^n}{1-r},r\neq 1$$ If $$|r|<1$$ we have $$r^n\to 0$$ and the above series converges to $$1/(1-r)$$.
For a typical use of Taylor (or Maclaurin) series one writes $$f(x) =\sum_{k=1}^{n} \frac{x^{k-1}}{(k-1)!}f^{(k-1)}(0)+R_n(x)$$ and there are ways to express remainder $$R_n(x)$$ in suitable forms which allow us to study the behavior of $$R_n(x)$$ as $$n\to\infty$$.
If for some range of values of $$x$$ we can prove that $$R_n(x) \to 0$$ then by definition (of convergence) it proves that the series $$\sum_{k=0}^{\infty}\frac{x^k}{k!}f^{(k)}(0)$$ is convergent and its sum is $$f(x)$$ for the values of $$x$$ in that range.