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.↩︎

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