Another very important expansion in powers of x is that for log(1+x). Since log(1+x)=0xdt1+t, and 1/(1+t)=1t+t2 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 1t+t2 from t=0 to t=x,  to the series x12x2+13x3. And this is in fact the case. For 1/(1+t)=1t+t2+(1)m1tm1+(1)mtm1+t, and so, if x>1, log(1+x)=0xdt1+t=xx22++(1)m1xmm+(1)mRm, where Rm=0xtmdt1+t.

We require to show that the limit of Rm, when m tends to , is zero. This is almost obvious when 0<x1; for then Rm is positive and less than 0xtmdt=xm+1m+1, and therefore less than 1/(m+1). If on the other hand 1<x<0, we put t=u and x=ξ, so that Rm=(1)m0ξumdu1u, which shows that Rm has the sign of (1)m. Also, since the greatest value of 1/(1u) in the range of integration is 1/(1ξ), we have 0<|Rm|<11ξ0ξumdu=ξm(m+1)(1ξ)<1(m+1)(1ξ): and so Rm0.

Hence log(1+x)=x12x2+13x3, provided that 1<x1. If x lies outside these limits the series is not convergent. If x=1 we obtain log2=112+13, a result already proved otherwise (Ex. LXXXIX. 7).


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

212. Series connected with the exponential and logarithmic functions. Expansion of ex by Taylor’s Theorem. Main Page 214. The series for the inverse tangent