Range-reduced exponential, logarithm and arctangent series have certified tails

For $k\ge N+1$, consecutive absolute exponential terms have ratio $\mathrm{\mid }u\mathrm{\mid }\mathrm{/}(k+1)\le q:=\mathrm{\mid }u\mathrm{\mid }\mathrm{/}(N+2)<1$. Their tail is bounded by the first omitted term times $1+q+q^2+\cdots =1\mathrm{/}(1-q)$, giving $R_E$. The defining power series lies between its partial sum plus or minus this absolute tail. For $N\ge 1$, the partial sum is at least $1-\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}2$ by alternating-series grouping when $u<0$ and by positivity when $u\ge 0$; also $R_E\le (1\mathrm{/}2{)}^2\mathrm{/}(2(1-1\mathrm{/}6))=3\mathrm{/}20<1\mathrm{/}2$. Thus the lower bound is positive.

For $y\in [1,2]$, $v=(y-1)\mathrm{/}(y+1)\in [0,1\mathrm{/}3]$. Set $G(v)=2{\sum }_{k=0}^{\mathrm{\infty }}{v}^{2k+1}\mathrm{/}(2k+1)$. Its derivative series is absolutely convergent for $\mathrm{\mid }v\mathrm{\mid }<1$ by [L6], so [L4] gives $G^{\mathrm{\prime }}(v)=2\mathrm{/}(1-v^2)$. By [L3], $H(y)=G(v(y))$ has derivative $4\mathrm{/}((y+1{)}^2(1-v^2))=1\mathrm{/}y$. The integral definition [L1] has derivative $1\mathrm{/}y$, since its difference quotient is the average of the continuous integrand $1\mathrm{/}t$ near $y$; also $H(1)=\ln 1=0$. For $y>1$, apply [L2] to $H-\ln$ on $[1,y]$; its derivative is zero, so $H(y)-\ln y=H(1)-\ln 1=0$. Equality at $y=1$ holds by the integral definition. Hence $H(y)=\ln y$ on $[1,2]$.

Fix $y>0$ and let $h(x)=\ln (xy)-\ln x$ for $x>0$. From [L1], the integral difference quotient gives $(\ln x{)}^{\mathrm{\prime }}=1\mathrm{/}x$. The chain rule [L3] therefore gives $h^{\mathrm{\prime }}(x)=y\mathrm{/}(xy)-1\mathrm{/}x=0$. Applying [L2] on the positive segment between $1$ and any $x>0$ shows $h(x)=h(1)=\ln y$, so $\ln (xy)=\ln x+\ln y$. Repeated use gives the identity for positive integer powers of $2$; applying it to $2\cdot (1\mathrm{/}2)=1$ gives $\ln (1\mathrm{/}2)=-\ln 2$, hence the same identity for negative integer powers and $j=0$. Thus $\ln (2^{j}y)=j\ln 2+\ln y$ for every integer $j$.

For $\mathrm{\mid }t\mathrm{\mid }\le 1\mathrm{/}2$, the power series $P(t)={\sum }_{k=0}^{\mathrm{\infty }}(-1{)}^k{t}^{2k+1}\mathrm{/}(2k+1)$ has radius $1$ by [L6], and [L4] gives $P^{\mathrm{\prime }}(t)={\sum }_{k\ge 0}(-1{)}^k{t}^{2k}=1\mathrm{/}(1+t^2)$ by the geometric series. The integral definition [L1] has the same derivative by the average-value difference quotient and the same value $0$ at $t=0$. If $t>0$, [L2] applied to $P-\arctan$ on $[0,t]$ gives equality; if $t<0$, apply [L2] on $[t,0]$; at $t=0$ both are zero. Thus $P(t)=\arctan t$ on the stated interval.

For each $k\ge N+1$, the absolute logarithm-series term is at most $2\mathrm{\mid }v{\mathrm{\mid }}^{2N+3}\mathrm{\mid }v{\mathrm{\mid }}^{2(k-N-1)}\mathrm{/}(2N+3)$. Summing this geometric majorant gives $2\mathrm{\mid }v{\mathrm{\mid }}^{2N+3}\mathrm{/}((2N+3)(1-v^2))$. Removing the first $N+1$ terms of $H$ leaves exactly this bounded tail.

For every omitted term index $k\ge N+1$, its magnitude is at most $\mathrm{\mid }t{\mathrm{\mid }}^{2N+3}\mathrm{\mid }t{\mathrm{\mid }}^{2(k-N-1)}\mathrm{/}(2N+3)$. Summing gives the displayed geometric tail, so $A_N(t)$ plus or minus that tail encloses $\arctan t$.