# 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.