213. The logarithmic series

Another very important expansion in powers of $x$ is that for $\log (1+x)$. Since $$\log (1+x)={\int }_0^x\frac{dt}{1+t},$$ and $1/(1+t)=1–t+t^2–\dots$ if $t$ is numerically less than unity, it is natural to expect¹ that $\log (1+x)$ will be equal, when $-1<x<1$, to the series obtained by integrating each term of the series $1–t+t^2–\dots$ from $t=0$ to $t=x$,  to the series $x–\frac{1}{2}{x}^2+\frac{1}{3}{x}^3–\dots$. And this is in fact the case. For $$1/(1+t)=1–t+t^2–\cdots +(-1{)}^{m-1}{t}^{m-1}+\frac{(-1{)}^m{t}^m}{1+t},$$ and so, if $x>-1$, $$\log (1+x)={\int }_0^x\frac{dt}{1+t}=x–\frac{x^2}{2}+\cdots +(-1{)}^{m-1}\frac{x^m}{m}+(-1{)}^m{R}_m,$$ where $$R_m={\int }_0^x\frac{t^{m}dt}{1+t}.$$

We require to show that the limit of $R_m$, when $m$ tends to $\mathrm{\infty }$, is zero. This is almost obvious when $0<x\le 1$; for then $R_m$ is positive and less than $${\int }_0^x{t}^{m}dt=\frac{x^{m+1}}{m+1},$$ and therefore less than $1/(m+1)$. If on the other hand $-1<x<0$, we put $t=-u$ and $x=-\xi$, so that $$R_m=(-1{)}^m{\int }_0^{\xi }\frac{u^{m}du}{1–u},$$ which shows that $R_m$ has the sign of $(-1{)}^m$. Also, since the greatest value of $1/(1–u)$ in the range of integration is $1/(1–\xi )$, we have $$0<|R_m|<\frac{1}{1–\xi }{\int }_0^{\xi }{u}^{m}du=\frac{{\xi }^m}{(m+1)(1–\xi )}<\frac{1}{(m+1)(1–\xi )}:$$ and so $R_m\to 0$.

Hence $$\log (1+x)=x–\frac{1}{2}{x}^2+\frac{1}{3}{x}^3–\dots ,$$ provided that $-1<x\le 1$. If $x$ lies outside these limits the series is not convergent. If $x=1$ we obtain $$\log 2=1–\frac{1}{2}+\frac{1}{3}–\dots ,$$ a result already proved otherwise (Ex. LXXXIX. 7).