Find the interval of convergence of the power series (-1^n)/(n)*(x-3)^(n-1). n=1.

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Find the interval of convergence of the power series (-1^n)/(n)*(x-3)^(n-1). n=1.

Mathematics
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\[\sum_{n=1}^{\infty} \frac{ (-1)^n }{ n(x-3)^{n-1} }\] is it this?
Sorry the denominator ends at n. The (x-3)^(n-1) is multiplied with the fraction
((-1^n)/(n))*(x-3)^(n-1)

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\[\sum_{n=1}^{\infty} \frac{ (-1)^n }{ n } (x-3)^{(n-1)}\]
we can try to apply the ratio test
That should work
\[|3-x|<1\]
more explicitly, we have the subsequent steps: \[\large \begin{gathered} \left| {{a_n}} \right| = \frac{{{{\left| {x - 3} \right|}^{n - 1}}}}{n} \hfill \\ \hfill \\ \frac{{\left| {{a_{n + 1}}} \right|}}{{\left| {{a_n}} \right|}} = \frac{{{{\left| {x - 3} \right|}^n}}}{{n + 1}} \cdot \frac{n}{{{{\left| {x - 3} \right|}^{n - 1}}}} = \frac{n}{{n + 1}}\left| {x - 3} \right| \to \left| {x - 3} \right| \hfill \\ \end{gathered} \]
now, please keep in mind if a series converges absolutely, then it converges in the ordinary meaning
absolutely convergent series behave nicely like finite sums; strictly conditionally convergent series do not
$$\sum_{n=1}^{\infty} \frac{ (-1)^n }{ n } (x-3)^{(n-1)}=-\sum_{n=0}^\infty \frac1{n+1}((-x)+3)^n$$ now consider that $$f(x)=\sum_{n=0}^\infty a_n x^n\\\int f\, dx=C+\sum_{n=0}^\infty \frac{a_n}{n+1}x^{n+1}\\\frac1x\int f\, dx=\frac{C}x+\sum_{n=0}^\infty\frac{a_n}{n+1}x^n$$ so we have that $$\sum_{n=0}^\infty\frac1{n+1}y^n=\frac1y \int\sum_{n=0}^\infty y^n\, dy=\frac1y\cdot\int\frac1{1-y}\, dy=-\frac1y\log(1-y)\\\implies -\sum_{n=0}^\infty \frac1{n+1}((-x)+3)^n=\frac1{3-x}\log (x-2)$$ about \(x=3\), which converges in a radius of \(1\) because of the singularity in \(\log(x-2)\) at \(x=2\)

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