## anonymous one year ago Find the interval of convergence of the power series (-1^n)/(n)*(x-3)^(n-1). n=1.

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1. Astrophysics

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

2. anonymous

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

3. anonymous

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

4. Astrophysics

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

5. Michele_Laino

we can try to apply the ratio test

6. Astrophysics

That should work

7. Astrophysics

$|3-x|<1$

8. Michele_Laino

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

9. Michele_Laino

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

10. anonymous

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

11. anonymous

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