Question

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

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^n }{ n(x-3)^{n-1} }\] is it this?

Answer 2

Sorry the denominator ends at n. The (x-3)^(n-1) is multiplied with the fraction

Answer 3

((-1^n)/(n))*(x-3)^(n-1)

Answer 4

\[\sum_{n=1}^{\infty} \frac{ (-1)^n }{ n } (x-3)^{(n-1)}\]

Answer 5

we can try to apply the ratio test

Answer 6

That should work

Answer 7

\[|3-x|<1\]

Answer 8

more explicitly, we have the subsequent steps:
\[\large \begin{gathered}
\left| {{a_n}} \right| = \frac{{{{\left| {x - 3} \right|}^{n - 1}}}}{n} \hfill \\
\hfill \\
\frac{{\left| {{a_{n + 1}}} \right|}}{{\left| {{a_n}} \right|}} = \frac{{{{\left| {x - 3} \right|}^n}}}{{n + 1}} \cdot \frac{n}{{{{\left| {x - 3} \right|}^{n - 1}}}} = \frac{n}{{n + 1}}\left| {x - 3} \right| \to \left| {x - 3} \right| \hfill \\
\end{gathered} \]

Answer 9

now, please keep in mind if a series converges absolutely, then it converges in the ordinary meaning

Answer 10

absolutely convergent series behave nicely like finite sums; strictly conditionally convergent series do not

Answer 11

$$\sum_{n=1}^{\infty} \frac{ (-1)^n }{ n } (x-3)^{(n-1)}=-\sum_{n=0}^\infty \frac1{n+1}((-x)+3)^n$$ 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 \(x=2\)