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lim-n-U-n-1-Un-gt-0-Test-for-convergence-




Question Number 85708 by Roland Mbunwe last updated on 24/Mar/20
lim_(n−∞)   /((U_n +1)/(Un))/   >0  Test for convergence
$${li}\underset{{n}−\infty} {{m}}\:\:/\frac{{U}_{{n}} +\mathrm{1}}{{Un}}/\:\:\:>\mathrm{0} \\ $$$${Test}\:{for}\:{convergence} \\ $$
Answered by Rio Michael last updated on 24/Mar/20
test for convergence states that   if U_n  is a sequence and   • lim_(x→∞)  ∣(U_(n+1) /U_n )∣ > 1  ⇒ U_n  is divergent.  • lim_(x→∞)  ∣(U_(n+1) /U_n )∣ < 1 ⇒ U_n  is convergent.  • lim_(x→∞)  ∣(U_(n+1) /U_n )∣ = 1 ⇒ inconclusive
$$\mathrm{test}\:\mathrm{for}\:\mathrm{convergence}\:\mathrm{states}\:\mathrm{that} \\ $$$$\:\mathrm{if}\:{U}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{and} \\ $$$$\:\bullet\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\mid\:>\:\mathrm{1}\:\:\Rightarrow\:{U}_{{n}} \:\mathrm{is}\:\mathrm{divergent}. \\ $$$$\bullet\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\mid\:<\:\mathrm{1}\:\Rightarrow\:{U}_{{n}} \:\mathrm{is}\:\mathrm{convergent}. \\ $$$$\bullet\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\mid\:=\:\mathrm{1}\:\Rightarrow\:\mathrm{inconclusive} \\ $$

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