= Logarithmic series from an integral Taylor remainder
{title2=$-\log(1-t)=\sum_{k=1}^{\infty}t^k/k,\quad 0\leq t<1$}
The <integral remainder in Taylor theorem> for $\log(1-t)$ gives the exact finite identity
$$
-\log(1-t)=\sum_{k=1}^N\frac{t^k}{k}+\int_0^t\frac{(t-x)^N}{(1-x)^{N+1}}\,dx.
$$
For $0\leq x\leq t<1$, $(t-x)/(1-x)\leq t$, so the nonnegative remainder is at most $t^{N+1}/(1-t)$ and tends to zero. At $t=1/2$ it is at most $2^{-N}$, proving both convergence to $\log2$ and an explicit error bound without assuming the logarithmic <power series> beforehand.
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