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For real functions $u,v$ that are continuously differentiable on a closed interval $[a,b]$, integration by parts statesIt follows from the product rule and the fundamental theorem of calculus. With the convention ${\int }_a^b=-{\int }_b^a$, the identity also applies to reversed limits. These regularity assumptions ensure that all the proper integrals exist.

For the smooth $f$ in the problem, define the integral remainder in Taylor theorem byThe fundamental theorem of calculus gives $f(t)=f(0)+R_1(t)$. Applying integration by parts to $R_n$, with $u=f^{(n)}(x)$ and antiderivative $v=-(t-x{)}^n/n!$ of $(t-x{)}^{n-1}/(n-1)!$, givesInduction starting from $n=1$ therefore provesThe same calculation holds for negative $t$: all integrals are oriented, and the segment joining $0$ and $t$ lies in $(-1,1)$. Thus the proof covers every stated $t$ and every positive integer $n$.

Now take $f(x)=\log (1-x)$. For $k\ge 1$, differentiation gives $f^{(k)}(x)=-(k-1)!/(1-x{)}^k$, and $f(0)=0$. With $n=N+1$ and $t=1/2$, the Taylor formula with integral remainder becomesFor $0\le x\le 1/2$, $0\le (1/2-x)/(1-x)\le 1/2$ and $(1-x{)}^{-1}\le 2$. Hence $0\le J_N\le 2^{-N}$ after integrating over an interval of length $1/2$. Thus the logarithmic series from an integral Taylor remainder satisfiesThe explicit vanishing remainder establishes convergence and identifies the sum, without using a pre-existing logarithmic power series.