= Solution
For real functions $u,v$ that are continuously differentiable on a closed interval $[a,b]$, <integration by parts> states
$$
\boxed{\int_a^b u(x)v'(x)\,dx=[u(x)v(x)]_a^b-\int_a^b u'(x)v(x)\,dx.}
$$
It 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> by
$$
R_n(t)=\frac1{(n-1)!}\int_0^t f^{(n)}(x)(t-x)^{n-1}\,dx.
$$
The <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 $(t-x)^{n-1}/(n-1)!$, gives
$$
R_n(t)=\frac{f^{(n)}(0)t^n}{n!}+\frac1{n!}\int_0^t f^{(n+1)}(x)(t-x)^n\,dx=\frac{f^{(n)}(0)t^n}{n!}+R_{n+1}(t).
$$
Induction starting from $n=1$ therefore proves
$$
\boxed{f(t)=\sum_{k=0}^{n-1}\frac{f^{(k)}(0)t^k}{k!}+\frac1{(n-1)!}\int_0^t f^{(n)}(x)(t-x)^{n-1}\,dx.}
$$
The 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 $k\ge1$, 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> becomes
$$
\log2=\sum_{k=1}^N\frac1{k2^k}+J_N,\qquad J_N=\int_0^{1/2}\frac{(1/2-x)^N}{(1-x)^{N+1}}\,dx.
$$
For $0\le x\le1/2$, $0\le(1/2-x)/(1-x)\le1/2$ and $(1-x)^{-1}\le2$. Hence $0\le J_N\le2^{-N}$ after integrating over an interval of length $1/2$. Thus the <logarithmic series from an integral Taylor remainder> satisfies
$$
\boxed{\sum_{k=1}^\infty\frac1{k2^k}=\log2,\qquad 0\le\log2-\sum_{k=1}^N\frac1{k2^k}\le2^{-N}.}
$$
The explicit vanishing remainder establishes convergence and identifies the sum, without using a pre-existing logarithmic power series.
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