Taylor Expansion of Natural Logarithm

golanor

That's why it is undetermined and not infinity.
x=50 is the smallest x for which the limit will always be 0.
anything lower than 50, and the nominator could be larger than the denominator => the limit will diverge.

mfb

"It could be larger" => you cannot use the general version of the lagrangian error to show convergence
It converges, but you cannot use this way to show it.

With c arbitrary small, it would diverge for other x as well, by the way.

golanor

yes, but since c is between x and 100, if x is bigger than 50 or smaller than 200 it won't diverge for any x and for any c.
when i pick x=1, for example, for 1<c<99 the limit diverges.
you cannot talk about a series in that case, because the remainder does not necessarily converge to 0.

mfb

Let me rewrite your limit a bit:

$$\lim_{n \to \infty} \frac{1}{c(n+1)} \frac{(x-100)^n}{c^n}$$

Consider c=100-epsilon with some (small) constant epsilon: The first part goes to 0, the second part, neglecting its sign, can be written as
$$\frac{(100-x)^n}{(100-\epsilon)^n}$$
This is smaller than 1 for epsilon<x, which is easy to satisfy for x<50 (c is close to 100 then). x=50, c=50 gives convergence as well, and you covered the remaining parts.

The series can converge to 0. Where is the problem?

golanor

It's very confusing. If the Lagrange remainder doesn't approach 0 it means you cannot estimate the function correctly, but in this case the calculation of the series does give accurate estimations.
It's very interesting.
Does anyone know the answer to this?

lurflurf

^The Lagrange remainder does approach 0, you have chosen a terrible way to estimate it. Either use a more subtle estimate of the Lagrange remainder or use the Cauchy Remainder or Schlömilch Remainder. Remember your estimate for c is quite crude.

golanor

The remainder must approach 0 for every c between x and 100.
If i can find even 1 such number for which the remainder does not approach 0, it means that the series is not a correct estimation for the function.

mfb

No: If you cannot find even 1 sequence c_n, ...
But I found one.