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Question Number 120810 by fajri last updated on 03/Nov/20
$${Determine}\:{the}\:{convergence}\:{intervval}\:{of}\:: \\ $$$$\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\left({x}\:−\:\mathrm{1}\right)^{{n}} \\ $$
Answered by 675480065 last updated on 03/Nov/20
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\mathrm{U}_{\mathrm{n}+\mathrm{1}} }{\mathrm{U}_{\mathrm{n}} }\mid<\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\mid\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}} }\mid<\mathrm{1} \\ $$$$\Rightarrow\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\mid\frac{−\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{x}−\mathrm{1}\right)}{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{n}} }\mid<\mathrm{1} \\ $$$$\Rightarrow\:\mid\mathrm{x}−\mathrm{1}\mid<\mathrm{2}\:\Rightarrow\:−\mathrm{2}<\mathrm{x}−\mathrm{1}<\mathrm{2} \\ $$$$\mathrm{hence}\:−\mathrm{1}<\mathrm{x}<\mathrm{3} \\ $$
Commented by fajri last updated on 03/Nov/20
$${thanks}\:{sir}... \\ $$