Question Number 185038 by Mastermind last updated on 16/Jan/23
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{for}\:\mathrm{which} \\ $$$$\mathrm{the}\:\mathrm{series}\:\frac{\mathrm{x}}{\mathrm{27}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{125}}+...+\frac{\mathrm{x}^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }+... \\ $$$$\mathrm{is}\:\mathrm{absolutely}\:\mathrm{convergent}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by FelipeLz last updated on 16/Jan/23
![S = Σ_(n=1) ^∞ (x^n /((2n+1)^3 )) lim_(n→∞) ∣((x^(n+1) /([2(n+1)+1]^3 ))/(x^n /((2n+1)^3 )))∣ < 1 lim_(n→∞) ∣x×(((2n+1)/(2n+3)))^3 ∣ < 1 lim_(n→∞) ∣x∣×lim_(n→∞) ∣(((2n+1)/(2n+3))×(n^(−1) /n^(−1) ))^3 ∣ < 1 ∣x∣×lim_(n→∞) ∣(((2+n^(−1) )/(2+3n^(−1) )))^3 ∣ < 1 ∣x∣ < 1](Q185048.png)
$${S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\frac{{x}^{{n}+\mathrm{1}} }{\left[\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}\right]^{\mathrm{3}} }}{\frac{{x}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }}\mid\:<\:\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid{x}×\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right)^{\mathrm{3}} \mid\:<\:\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid{x}\mid×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}×\frac{{n}^{−\mathrm{1}} }{{n}^{−\mathrm{1}} }\right)^{\mathrm{3}} \mid\:<\:\mathrm{1} \\ $$$$\mid{x}\mid×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid\left(\frac{\mathrm{2}+{n}^{−\mathrm{1}} }{\mathrm{2}+\mathrm{3}{n}^{−\mathrm{1}} }\right)^{\mathrm{3}} \mid\:<\:\mathrm{1} \\ $$$$\mid{x}\mid\:<\:\mathrm{1} \\ $$