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I am really stuck with this. I got a taylor series of ${((-1)^{(n+1)}z^n)}/{(21^n(n!))}$ at $z=0$ if $n \neq 0$; however, wolfram alpha gave me a different answer as attached. enter image description here

My intuition tells me that the taylor series converges in the circle $|z| < 21$ as we cannot have $z \leq -21$; however, when I apply ratio test. I obtain no $n's$ in the numerator and one $n$ in the denominator so the limit at infinity would be 0 regardless of the value of $z$ so I am led to believe that it converges for all $z$, is this correct?

Please provide as detailed of an explanation as possible especially with the ratio test.

Aaron Hendrickson
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Elli
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    The, $n!$ in the denominator is wrong. Do you know the Taylor series for $\log(1+z)?$ Because $$\log(z+21)=\log(21)+\log\left(1+\frac z{21}\right)$$ – Thomas Andrews Jun 17 '21 at 02:06
  • Honestly, I'm not too sure. Can you show me how to obtain the taylor series for this example. It seems that is incorrect – Elli Jun 17 '21 at 03:51
  • Also, how did you get that expressison for $log(z+21)$ – Elli Jun 17 '21 at 04:07

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