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 $1/(1 + t) = 1 – t + t^{2} – \dots + (-1)^{m-1} t^{m-1} + \frac{(-1)^{m} t^{m}}{1 + t},$ and so, if $$x > -1$$, $\log(1 + x) = \int_{0}^{x} \frac{dt}{1 + t} = x – \frac{x^{2}}{2} + \dots + (-1)^{m-1} \frac{x^{m}}{m} + (-1)^{m} R_{m},$ 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 $$\infty$$, is zero. This is almost obvious when $$0 < x \leq 1$$; 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 $\log(1 + x) = x – \tfrac{1}{2} x^{2} + \tfrac{1}{3} x^{3} – \dots,$ provided that $$-1 < x \leq 1$$. If $$x$$ lies outside these limits the series is not convergent. If $$x = 1$$ we obtain $\log 2 = 1 – \tfrac{1}{2} + \tfrac{1}{3} – \dots,$ a result already proved otherwise (Ex. LXXXIX. 7).

1. See Appendix II for some further remarks on this subject.↩︎