Matrix Operator Method

[bo198214]

Hm, then I somewhere must have an error in my computation.
I wanted to compute the Matrix logarithm via the formula
\( \log(A)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (A-I)^n \).
For A being the power derivation matrix of \( e^x \) at 0, i.e. truncated to 6x6
\(
A=\left( \begin{matrix}
1&0&0&0&0&0\\
1&1&1/2&1/6&1/24&{\frac {1}{120}}\\
1&2&2&4/3&2/3&{\frac {4}{
15}}\\
1&3&9/2&9/2&{\frac {27}{8}}&{\frac {81}{40}}
\\
1&4&8&{\frac {32}{3}}&{\frac {32}{3}}&{\frac {128}
{15}}\\
1&5&{\frac {25}{2}}&{\frac {125}{6}}&{\frac {
625}{24}}&{\frac {625}{24}}
\end{matrix} \right)
\)
(which is transposed to the matrix you usually uses).

This Matrix is diagonalizable, but if I compute the logarithm via the above sum then the entries do not converge. For example:
\( \sum_{n=1}^{10} \frac{(-1)^{n+1}}{n} (A-I)^n\\=
\left( \begin{matrix}
0.0& 0.0& 0.0& 0.0& 0.0& 0.0\\
- 44101769800.0&- 145224229100.0&- 279795780200.0&
- 388116737400.0&- 423988471200.0&- 383041240500.0\\
- 594655919700.0&- 1958162913000.0&- 3772688135000.0&- 5233257693000.0
&- 5716942159000.0&- 5164821148000.0\\
- 3307574479000.0&- 10891625760000.0&- 20984315030000.0&-
29108244400000.0&- 31798577400000.0&- 28727589090000.0
\\
- 11906282180000.0&- 39206606270000.0&-
75537279400000.0&- 104781003800000.0&- 114465400800000.0&-
103410758300000.0\\
- 33057719360000.0&-
108856901800000.0&- 209728793100000.0&- 290923814800000.0&-
317812483800000.0&- 287119336600000.0
\end{matrix} \right)
\)