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anonymous
 one year ago
Find the interval of convergence of the power series (1^n)/(n)*(x3)^(n1). n=1.
anonymous
 one year ago
Find the interval of convergence of the power series (1^n)/(n)*(x3)^(n1). n=1.

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{n=1}^{\infty} \frac{ (1)^n }{ n(x3)^{n1} }\] is it this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry the denominator ends at n. The (x3)^(n1) is multiplied with the fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0((1^n)/(n))*(x3)^(n1)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{n=1}^{\infty} \frac{ (1)^n }{ n } (x3)^{(n1)}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can try to apply the ratio test

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more explicitly, we have the subsequent steps: \[\large \begin{gathered} \left {{a_n}} \right = \frac{{{{\left {x  3} \right}^{n  1}}}}{n} \hfill \\ \hfill \\ \frac{{\left {{a_{n + 1}}} \right}}{{\left {{a_n}} \right}} = \frac{{{{\left {x  3} \right}^n}}}{{n + 1}} \cdot \frac{n}{{{{\left {x  3} \right}^{n  1}}}} = \frac{n}{{n + 1}}\left {x  3} \right \to \left {x  3} \right \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, please keep in mind if a series converges absolutely, then it converges in the ordinary meaning

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0absolutely convergent series behave nicely like finite sums; strictly conditionally convergent series do not

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\sum_{n=1}^{\infty} \frac{ (1)^n }{ n } (x3)^{(n1)}=\sum_{n=0}^\infty \frac1{n+1}((x)+3)^n$$ 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 $$\sum_{n=0}^\infty\frac1{n+1}y^n=\frac1y \int\sum_{n=0}^\infty y^n\, dy=\frac1y\cdot\int\frac1{1y}\, dy=\frac1y\log(1y)\\\implies \sum_{n=0}^\infty \frac1{n+1}((x)+3)^n=\frac1{3x}\log (x2)$$ about \(x=3\), which converges in a radius of \(1\) because of the singularity in \(\log(x2)\) at \(x=2\)
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