I have some doubts concerning the proof of the "term-wise differentiation of power series" theorem. Below, I first included 3 theorems that are used in the proof; then, I included the whole proof and state the related theorems; finally, I included my specific doubts about it.
The theorems used in the proof are basically the Squeeze Theorem and the facts that the first and second derivatives of a power series have the same radius of convergence as the original power series. They are stated as follows:
Theorem 1: If the radius of convergence of the power series $\sum_{n=0}^{+\infty} c_nx^n$ is $R>0$, then R is also the radius of convergence of the series $\sum_{n=1}^{+\infty} nc_nx^{n-1}.$
Theorem 2: If the radius of convergence of the power series $\sum_{n=0}^{+\infty} c_nx^n$ is $R>0$, then R is also the radius of convergence of the series $\sum_{n=2}^{+\infty} n(n-1)c_nx^{n-2}.$
Theorem 3 (also known as Squeeze Theorem): Assume $f$, $g$ and $h$ are functions defined in an open interval $I$ containing $a$, except, possibly, at $a$ itself, and that $f(x) \leq g(x) \leq h(x)$ for all $x$ in $I$ such that $x\neq a$. If both $\lim_{x\to a}f(x)$ and $\lim_{x\to a}h(x)$ exist and are equal to $L$, then $\lim_{x\to a}g(x)$ also exists and is equal to $L$.
The theorem statement and its proof, about which I have some questions, are below:
Theorem: Let $\sum_{n=0}^{+\infty} c_nx^n$ be a power series whose radius of convergence is $R>0$. Then, if $f$ is the function defined by
$$f(x) = \sum_{n=0}^{+\infty} c_nx^n \ \ \ \ (1)$$
$f'(x)$ exists for all $x$ in the open interval $(-R,R)$, and it is given by
$$f(x) = \sum_{n=1}^{+\infty} n c_nx^{n-1}$$
PROOF: Let $x$ and $a$ be two distinct numbers in the open interval $(-R,R)$. The Taylor formula, with $n = 1$, is
$$f(x) = f(a) + \dfrac{f'(a)}{1!}(x-a) + \dfrac{f''(\xi)}{2!}(x-a)^2$$
Using this formula with $f(x) = x^n$, we have, for every positive integer $n$,
$$x^n=a^n+na^{n-1}(x-a)+\frac{1}{2}n(n-1)(\xi_n)^{n-2}(x-a)^2 \ \ \ \ (2)$$
where $\xi_n$ is between $a$ and $x$, for every positive integer $n$. From (1) we have
$$\begin{align}
f(x)-f(a) &= \sum_{n=0}^{+\infty} c_nx^n – \sum_{n=0}^{+\infty} c_na^n\\
&= c_0 + \sum_{n=1}^{+\infty} c_nx^n – c_0 – \sum_{n=1}^{+\infty} c_na^n\\
&= \sum_{n=1}^{+\infty} c_n(x^n – a^n)
\end{align}$$
Dividing by $x-a$ (because $x\neq a$) and using (2), we have, from the above equation
$$ \dfrac{f(x)-f(a)}{x-a} = \dfrac{1}{x-a} \sum_{n=1}^{+\infty} c_n[na^{n-1}(x-a)+\frac{1}{2}n(n-1)(\xi_n)^{n-2}(x-a)^2] $$
Thus,
$$ \dfrac{f(x)-f(a)}{x-a} = \sum_{n=1}^{+\infty} nc_na^{n-1}+\frac{1}{2}(x-a)\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2} \ \ \ \ (3)$$Since $a$ is in $(-R,R)$, it follows from Theorem 1 that $\sum_{n=1}^{+\infty}nc_na^{n-1}$ is absolutely convergent.
Since both $a$ and $x$ are in $(-R,R)$, there exists a number $K > 0$ such that $|a|<K<R$ and $|x|<K<R$. It follows from Theorem 2 that
$$\sum_{n=2}^{+\infty}n(n-1)c_nK^{n-2}$$
is absolutely convergent. Then, since
$$|n(n-1)c_n(\xi_n)^{n-2}| < |n(n-1)c_nK^{n-2}| \ \ \ \ (4)$$
for every $\xi_n$, we can conclude, from the comparison test, that
$$\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2}$$
is absolutely convergent.It follows from (3) that
$$ \left|\dfrac{f(x)-f(a)}{x-a} – \sum_{n=1}^{+\infty} nc_na^{n-1}\right| = \left|\frac{1}{2}(x-a)\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2}\right| \ \ \ \ (5)$$
However, we know that, if $\sum_{n=1}^{+\infty}u_n$ is absolutely convergent, then
$$\left|\sum_{n=1}^{+\infty}u_n\right| \leq \sum_{n=1}^{+\infty}|u_n|$$
Applying this to the right side of (5), we obtain:
$$ \left|\dfrac{f(x)-f(a)}{x-a} – \sum_{n=1}^{+\infty} nc_na^{n-1}\right| \leq \frac{1}{2}|x-a|\sum_{n=2}^{+\infty}n(n-1)|c_n||\xi_n|^{n-2} \ \ \ \ (6)$$
From (4) and (6), we obtain:
$$ \left|\dfrac{f(x)-f(a)}{x-a} – \sum_{n=1}^{+\infty} nc_na^{n-1}\right| \leq \frac{1}{2}|x-a|\sum_{n=2}^{+\infty}n(n-1)|c_n|K^{n-2} \ \ \ \ (7)$$
where $0<K<R$. Since the series on the right side of (7) is absolutely convergent, the limit on the right side, as $x$ approaches $a$, is zero. Thus, from (17) and from Theorem 3 (Squeeze Theorem), it follows that
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\sum_{n=1}^{+\infty}nc_na^{n-1}$$
which is equivalent to
$$f'(a)=\sum_{n=1}^{+\infty}nc_na^{n-1}$$
and, since $a$ can be any number in the open interval $(-R,R)$, the theorem is proven.
My 2 doubts are:
- Why is the step that transforms (6) into (7) necessary? Assuming that the goal is to have the right side of the inequality to be convergent, then, since $\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2}$ is known to be absolutely convergent, isn't this goal already achieved by the summation on the right side of (6)?
- How is the last limit obtained from (7) and the Squeeze Theorem? It seems that, from (7) and the Squeeze Theorem, it follows that $\lim_{x\to a} \left[ \dfrac{f(x)-f(a)}{x-a} – \sum_{n=1}^{+\infty} nc_na^{n-1}\right] =0$. So, I think that, since $f(x)$ is differentiable at $x = a$, then I know that $lim_{x\to a} [f(x)-f(a)]/(x-a)$ exists; so I can simply distribute the limit over the sum and arrive at the result. Is this correct?
Best Answer
For question 1, I think that you are correct. There is no need to transform (6) into (7) because the writer has already observed that the series on the right-hand side of (6) is convergent.
For question 2, the application of the Squeeze Theorem that they have in mind is likely the following. If $C$ is the sum of the series on the right-hand side of (7), we can write the inequality (7) as
$$ \sum_{n=1}^\infty n c_n a^{n-1} - \frac{1}{2} |x-a| C \leq \frac{f(x) - f(a)}{x - a} \leq \sum_{n=1}^\infty n c_n a^{n-1} + \frac{1}{2} |x-a|C. $$
Letting $x \to a$, both the left and right sides of this double-inequality tend to $\sum_{n=1}^\infty n c_n a^{n-1}$, so the limit of the central expression must also exist and have the same value.