Suppose $0\leq x<1$, the Taylor expansion of the $(1+x)^\lambda$ at $x_0=0$ should be convergent. I want to prove this by using when $n\to \infty$ the remainder of this expansion is approaching zero without using other convergence test
$$ (1+x)^\lambda=1+\lambda x+\dfrac{\lambda(\lambda-1)}{2!}x^2+\cdots + \dfrac{\lambda (\lambda-1)\cdots(\lambda-(n-1))}{n!}x^n +r_n(x)$$
The remainder can be expressed as the Lagrange form $$ r_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$$
where $\xi \in (0,x)$
and $$ f^{(n+1)}(\xi) = \lambda (\lambda-1)\cdots(\lambda-(n-1))(\lambda-n)(1+\xi)^{\lambda-(n+1)} $$
Thus the remainder is $$ \dfrac{\lambda (\lambda-1)\cdots(\lambda-(n-1))(\lambda-n)(1+\xi)^{\lambda-(n+1)}}{(n+1)!}x^{n+1}$$
Then I simplify it as
$$ \dfrac{\lambda (\lambda-1)\cdots(\lambda-(n-1))(\lambda-n)}{(n+1)!}(1+\xi)^{\lambda}(
\frac{x}{1+\xi})^{n+1}$$.
And $n\to \infty$ it's clear to see that since $1+\xi>x$, $\ \ (\dfrac{x}{1+\xi})^{n+1}$ this term will go zero. Also $(1+\xi)^\lambda$ isn't a zero or infinity term
However, I don't know how to deal with $ \dfrac{\lambda (\lambda-1)\cdots(\lambda-(n-1))(\lambda-n)}{(n+1)!}=\displaystyle \binom{\lambda}{n+1}$ as $n \to \infty$.
Thus, any help on this? Or maybe there exists some other ways to deal with the remainder?
Best Answer
Just use the alternating series theorem with its "estimation" saying that if the general term of the series, here:
$$a_n=\dfrac{\lambda (\lambda-1)\cdots(\lambda-(n-1))}{n!}x^n\tag{1}$$
(i) tends to $0$ in a decreasing way in absolute value.
(ii) with alternate signs ($sign(a_{n+1})=-sign(a_{n}$),
then the series is convergent with
$$|S_n-S| \le |a_{n+1}|$$
In words: the error, i.e., the absolute value of the difference between the partial sum $S_n:=a_1+a_2+\cdots a_n$ and the total sum $S$ is bounded by the first "neglected term".
Here, in order to have the two conditions (i) and (ii) fulfilled, we need to drop a certain finite number of terms (as we are going to explain) which is not harmful.
Let us use the following consequence of (1):
$$a_{n} \ = \ a_n \ x \ \underbrace{\frac{(\lambda-(n-1))}{n}}_{K_n}$$
$$|a_n|<|a_{n-1}|x$$
As $x<1$, the latter relationship proves that $|a_n| \to 0$ when $n \to \infty$.
$$\lambda -(n-1) \le n \ \iff \ 2n-1 \ge \lambda$$
Otherwise said when:
$$ \ n \ge n_1:=\lfloor\frac{\lambda+1}{2}+1)\rfloor $$
Therefore, if $n \ge \max(n_0,n_1)$, the upsaid theorem can be applied.