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By the Taylor series formula (taking ),
where
First, .
Next, if we take a couple derivatives,
![{\displaystyle {\begin{aligned}f^{(1)}(x)={\frac {1}{x}},\quad f^{(2)}(x)=-{\frac {1}{x^{2}}},\quad f^{(3)}(x)={\frac {2}{x^{3}}},\quad f^{(4)}(x)=-{\frac {6}{x^{4}}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d51f3973467809be8a9db040409ab77cee609e3a)
we can find the following patterns: the sign keeps changing, the power of in denominator of is , and the number in numerator of is . Thus, we can get a general formula for :
![{\displaystyle {\begin{aligned}f^{(n)}(x)={\frac {(-1)^{n+1}(n-1)!}{x^{n}}},\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9875a1a86b8c49390ac11b6d9aa79b9a03efb80e)
which implies
![{\displaystyle {\begin{aligned}f^{(n)}(2)={\frac {(-1)^{n+1}(n-1)!}{2^{n}}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/268b1019636bd939e3784b19063387cba3303b4c)
and
![{\displaystyle {\begin{aligned}c_{n}={\frac {(-1)^{n+1}(n-1)!}{2^{n}n!}}={\frac {(-1)^{n+1}}{n2^{n}}}.\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/88ebd328a89c43426609ea267d95a5e8b6e01165)
Thus, we have the Taylor series:
![{\displaystyle {\begin{aligned}\ln x=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n2^{n}}}(x-2)^{n}.\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/220ea3252ba8bb40cce92c58749e4ea583da7e34)
For the interval of convergence, we will use the ratio test. First, compute
The series converges when this ratio is smaller than 1 and so for in , the series converges and for in the series diverges. We still have to check the endpoints, that is when . This occurs at and . At , the series becomes,
![{\displaystyle \ln x=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n2^{n}}}(-2)^{n}=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}(-1)^{n}}{n}}=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)}{n}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/835a3a8bedc9fff5281afd19776174e3fd718f76)
We know that this series diverges and therefore the Taylor series does not converge at . Note that we expect this to happen because . At , the series becomes,
![{\displaystyle \ln x=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n2^{n}}}(2)^{n}=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d7c83cbea9cd955ab41640eadf5c01f24ae43615)
This is an alternating series with decreasing terms that tend to zero and therefore by the alternating series test, converges. Once again, this makes sense since has a finite value. Combining everything together, the interval of convergence for the series is .
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