Taylor series for the logarithm

We first derive the Taylor series for f (x) = ln(x) with center a = 1.
By induction it is easy to prove that for every natural number k, the k-th derivative is f ^(k)(x) = (−1)^k+1(k − 1)!/x^k, hence the coefficients are

We had to (and will have to) treat the case k = 0 separately, since f ⁽⁰⁾ = f does not fit the general pattern. The Taylor series is

Now take any number x from the interval (0,2). We will use the Lagrange estimate of the remainder to prove that this Taylor series converges to the logarithm at this x. If we denote by I the closed interval with endpoints 1 and x, we have the estimate

Now we need to ask where x is. If x = 1, then T(1) = 1 = ln(1). If x > 1, then the interval I is [1,x], and since 1/t^N+1 is a decreasing function, the maximum happens at 1. Therefore

We used the fact that x < 2, therefore (x − 1) < 1 and we know that q^N goes to zero if |q| < 1 (see the geometric sequence). So for x > 1 (and of course for x < 2) we have the convergence of T(x) to ln(x) proven.

When x < 1, then the interval I is [x,1] and the maximum of the decreasing function 1/t^N+1 is attained at x. Therefore

Now we have a little problem. If x > 1/2, then the number in absolute value is less than 1 and we get convergence using the good old geometric sequence argument. On the other hand, for x between 0 and 1/2 this argument fails, since the number in the geometric sequence is larger than 1 and the upper Lagrange estimate goes to infinity. However, note that it is only an upper estimate, so this does not necessarily mean that the difference |ln(x) − T_N| goes to infinity. Indeed it does not, we know that also there it actually goes to zero, since in Theory - Series of functions - Taylor series we proved using a different method that the above Taylor series converges to logarithm on (0,2]. This shows that on (0,1/2), the Lagrange estimate is too generous.