Series and Power Series

For $M>N$ the partial-sum increment is bounded by $\abs{s_M-s_N}=\abs{\sum_{n=N+1}^M a_n}\le\sum_{n=N+1}^M \abs{a_n}$, which is the corresponding increment of the partial sums of $\sum_n\abs{a_n}$. Since the latter series converges its increments tend to $0$, so $(s_N)$ is Cauchy and converges by completeness. The inequality follows from $\abs{s_N}\le\sum_{n=1}^N\abs{a_n}$ by passing to the limit, using continuity of the absolute value.

If $L<1$, choose $r$ with $L<r<1$. By definition of the limit superior, $\abs{a_n}^{1/n}\le r$ for all large $n$, so $\abs{a_n}\le r^n$ eventually, and $\sum_n r^n$ converges as a geometric series with ratio below $1$, so $\sum_n\abs{a_n}$ converges by comparison. If $L>1$, then $\abs{a_n}^{1/n}\ge 1$ for infinitely many $n$, so $\abs{a_n}\ge 1$ infinitely often, the terms do not tend to $0$, so the series cannot converge.

Apply the root test to the series $\sum_n a_n x^n$, whose terms have $\limsup_n\abs{a_n x^n}^{1/n}=\abs x\,\limsup_n\abs{a_n}^{1/n}=\abs x/R$. This is below $1$ exactly when $\abs x<R$, giving absolute convergence there, and above $1$ when $\abs x>R$, giving divergence. When the limit superior is $\infty$, so $R=0$, the computed value is $\infty>1$ for every $x\neq 0$, leaving only $x=0$ convergent, and when it is $0$, so $R=\infty$, the value is $0<1$ for every $x$. The boundary $\abs x=R$ is left undecided, as it must be, since the behaviour there depends on the series.

For $x\neq 0$ the differentiated series is $\sum_{n\ge1} n a_n x^{n-1}=x^{-1}\sum_{n\ge1} n a_n x^n$, and multiplication by the fixed nonzero factor $x^{-1}$ does not affect convergence, so $\sum_n n a_n x^{n-1}$ and $\sum_n n a_n x^n$ converge for exactly the same $x$ (both at $x=0$ trivially) and share a radius. That radius is governed by $\limsup_n(n\abs{a_n})^{1/n}=\big(\lim_n n^{1/n}\big)\limsup_n\abs{a_n}^{1/n}= \limsup_n\abs{a_n}^{1/n}$, using $n^{1/n}\to 1$, which holds because $\log(n^{1/n})=(\log n)/n\to 0$. A convergent factor passes through the limit superior, so the two series share the value of $\limsup_n \abs{a_n}^{1/n}$ and hence the radius $R$.

Fix a point $a$ and form the difference quotients $\varphi_N(x)=\dfrac{g_N(x)-g_N(a)}{x-a}$ for $x\neq a$. For two indices the mean value theorem applied to $g_N-g_M$ gives a point $\xi$ with $\varphi_N(x)-\varphi_M(x)=g_N'(\xi)-g_M'(\xi)$, so $\abs{\varphi_N(x)-\varphi_M(x)}\le\sup\abs{ g_N'-g_M'}$, a bound independent of $x$. Since $g_N'\to h$ uniformly the right side tends to $0$, so $(\varphi_N)$ is uniformly Cauchy and converges uniformly to $\varphi(x)=\frac{f(x)-f(a)}{x-a}$. Each $\varphi_N$ has the limit $g_N'(a)$ as $x\to a$, so the iterated-limit (Moore-Osgood) theorem applies, its two hypotheses being the uniform convergence $\varphi_N\to\varphi$ just shown and the existence of each inner limit $\lim_{x\to a}\varphi_N(x)=g_N'(a)$. With $\lim_N g_N'(a)=h(a)$ existing by hypothesis it permits exchanging that limit with the limit in $N$, so

The left side is by definition $f'(a)$, so $f'(a)=h(a)$.

Fix $r<R$ and work on $\abs x\le r$. The partial sums $s_N(x)=\sum_{n\le N}a_n x^n$ are polynomials with $s_N'(x)=\sum_{n\le N}n a_n x^{n-1}$, and these derivatives converge uniformly on $\abs x\le r$, since $\abs{n a_n x^{n-1}}\le n\abs{a_n}r^{n-1}$ and $\sum_n n\abs{a_n}r^{n-1}$ converges by Lemma 5 and absolute convergence at $r<R$. By Theorem 6, $f$ is differentiable on $\abs x<r$ with $f'=\sum_n n a_n x^{n-1}$, and since $r<R$ was arbitrary this holds on $\abs x<R$. By Lemma 5 the derivative $f'=\sum_{n\ge1} n a_n x^{n-1}$ is itself a power series of radius $R$, so the argument reapplies to $f'$ and, by induction, $f^{(k)}$ exists as a power series of radius $R$ for every $k$. Then $f^{(k)}(x)=\sum_{n\ge k} n(n-1)\cdots(n-k+1)a_n x^{n-k}$, and evaluating the $k$-th derivative at $x=0$ leaves only the constant term $k!\,a_k$, so $a_k=f^{(k)}(0)/k!$, the Taylor coefficients.

Differentiating term by term with Corollary 7, $\exp'(x)=\sum_{n\ge 1}\frac{n x^{n-1}}{n!} =\sum_{n\ge 1}\frac{x^{n-1}}{(n-1)!}=\exp(x)$, and $\exp(0)=1$ from the constant term. For the functional equation, fix $y$ and let $u(x)=\exp(x+y)\exp(-x)$, whose derivative is $u'(x)=\exp(x+y)\exp(-x)-\exp(x+y)\exp(-x)=0$ by the product and chain rules with $\exp'=\exp$, so $u$ is constant by the mean value theorem, equal to its value $u(0)=\exp(y)$. Thus $\exp(x+y)\exp(-x)=\exp(y)$, and taking $y=0$ shows $\exp(-x)=1/\exp(x)$, so $\exp(x+y)=\exp(x)\exp(y)$.