Elementary factorial and geometric series converge uniformly

Let $0\le q<1$. For every integer $m\ge 0$, $(1-q){\sum }_{k=0}^m{q}^k=1-q^{m+1}$. If $q=0$, its powers after the zeroth are zero. If $0<q<1$, then $1\mathrm{/}q=1+d$ for $d>0$, and induction gives $(1+d{)}^m\ge 1+md$. By [L1], $1+md$ is unbounded, so $0\le q^m\le 1\mathrm{/}(1+md)\to 0$. Thus the geometric partial sums converge to $1\mathrm{/}(1-q)$, with tail $q^{m+1}\mathrm{/}(1-q)$.

For $R=0$ the factorial series is $1$. For $R>0$, choose $N$ with $q=R\mathrm{/}(N+2)<1$ by [L1]. If $u_n=R^n\mathrm{/}n!$, then for $n\ge N+1$, $u_{n+1}\mathrm{/}{u}_n=R\mathrm{/}(n+1)\le q$. Every finite tail after $N$ is therefore at most $u_{N+1}{\sum }_{j\ge 0}{q}^j=u_{N+1}\mathrm{/}(1-q)$ by step 1.1. The positive partial sums are increasing and bounded, so [F1] gives their supremum $S$. For every $\epsilon >0$, some partial sum exceeds $S-\epsilon$, since otherwise $S-\epsilon$ would be an upper bound; monotonicity then puts all later partial sums between $S-\epsilon$ and $S$. The finite-tail geometric bound passes to this limit and proves the stated estimate.

For $\mathrm{\mid }x\mathrm{\mid }\le R$, [L2] gives $\mathrm{\mid }{x}^n\mathrm{/}n!\mathrm{\mid }\le R^n\mathrm{/}n!$. Thus the exponential series is uniformly Cauchy by the tail bound in step 2.1. The sine and cosine series are subseries of this same nonnegative majorant, so their absolute tails are uniform as well. For the formal derivative series, the exponential terms are bounded by $R^{n-1}\mathrm{/}(n-1)!$; sine derivatives are bounded by $R^{2k}\mathrm{/}(2k)!$, and nonconstant cosine derivatives by $R^{2k-1}\mathrm{/}(2k-1)!$. Each is a subseries of a factorial series from step 2.1. The same tail comparison proves absolute and uniform convergence of all stated series.

For $0\le r<1$, the geometric series is uniformly absolutely convergent on $\mathrm{\mid }x\mathrm{\mid }\le r$ by step 1.1. Its formal derivative terms are bounded by $v_n=n{r}^{n-1}$ for $n\ge 1$. If $r=0$, only the first derivative term is nonzero. If $0<r<1$, choose $q$ with $r<q<1$ and an integer $n_0$ large enough that $r(1+1\mathrm{/}n)\le q$ for $n\ge n_0$, using [L1]. Then $v_{n+1}\mathrm{/}{v}_n=r(1+1\mathrm{/}n)\le q$, so the derivative majorant has tails bounded by a constant times geometric tails from step 1.1. This proves absolute and uniform convergence of the geometric derivative series. The absolute terms of ${\sum }_{k\ge 0}{x}^{2k+1}\mathrm{/}(2k+1)$ are bounded by $r^{2k+1}$ and the derivative terms by $r^{2k}$. Their tails converge uniformly by step 1.1 applied to $q=r^2$; multiplication by the fixed factor $r$ or $2$ does not affect convergence. This proves the assertions for the odd power series and its formal derivative series.