Why if $x <1$ then, $\sum_{n=0}^{\infty}(n+2)x^{n}<\infty$
I know that $\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}<\infty$
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So multiply the entire series by $x$ and then integrate term by term. – Troposphere Aug 18 '21 at 18:10
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1Do you know how to find the radius of convergence of a power series? – Mark Aug 18 '21 at 18:11
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For $$;x=-3<1;,;;\sum_{n=0}^\infty (n+2)x^n;;\text{does not converge}\ldots$$ So I am guessing you didn't actually mean "for$;x<1;$", but in fact... – DonAntonio Aug 18 '21 at 18:33
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How did you prove that $\sum x^n$ converges when $|x| < 1$? Similar techniques will work here. There are more powerful tools, such as the ratio text, but if you knew those, I don;t think you would be asking the question. – Doug M Aug 18 '21 at 18:38
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False. The series diverges if $x\le-1.$ – zhw. Aug 18 '21 at 20:30