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Question Number 122152 by liberty last updated on 14/Nov/20
(1) lim_(n→∞) ((1/( (√(n^2 +1))))+(1/( (√(n^2 +2))))+(1/( (√(n^2 +3))))+...+(1/( (√(n^2 +n+1)))) ) =? (2) lim_(n→∞) (((2 (n)^(1/n) −1)^n )/n^2 ) =?
$$\:\left(\mathrm{1}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{3}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}}}\:\right)\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}\:\sqrt[{\mathrm{n}}]{\mathrm{n}}\:−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:=? \\ $$
Answered by benjo_mathlover last updated on 14/Nov/20
(1) lim_(n→∞) ((1/( (√(n^2 +1))))+(1/( (√(n^2 +2))))+(1/( (√(n^2 +3))))+...+(1/( (√(n^2 +n+1))))) note that ((n+1)/( (√(n^2 +n+1)))) ≤ (1/( (√(n^2 +1))))+(1/( (√(n^2 +2))))+...+(1/( (√(n^2 +n+1)))) ≤ ((n+1)/( (√(n^2 +1)))) This and the squeeze law for sequences imply that the limit is 1. lim_(n→∞) ((n(1+(1/n)))/(n (√(1+(1/n)+(1/n^2 ))))) = lim_(n→∞) ((n(1+(1/n)))/(n (√(1+(1/n^2 ))))) = 1.
$$\left(\mathrm{1}\right)\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{3}}}+…+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{n}+\mathrm{1}}}\right) \\ $$$${note}\:{that}\: \\ $$$$\frac{{n}+\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{n}+\mathrm{1}}}\:\leqslant\:\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{2}}}+…+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{n}+\mathrm{1}}}\:\leqslant\:\frac{{n}+\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${This}\:{and}\:{the}\:{squeeze}\:{law}\:{for}\:{sequences} \\ $$$${imply}\:{that}\:{the}\:{limit}\:{is}\:\mathrm{1}. \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}{{n}\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}{{n}\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}}\:=\:\mathrm{1}. \\ $$
Commented by liberty last updated on 14/Nov/20
thanks
$$\mathrm{thanks} \\ $$
Answered by mathmax by abdo last updated on 14/Nov/20
U_n =(((2 n^(1/n) −1)^n )/n^2 ) ⇒ U_n =(((2n^(1/n) −1)^n )/((n^(2/n) )^n )) =(((2n^(1/n) −1)/n^(2/n) ))^n =(2 n^(−(1/n)) −n^(−(2/n)) )^n ⇒U_n =e^(nlog(2 n^(−(1/n)) −n^((−2)/n) )) v_n =nlog(2.n^(−(1/n)) −n^(−(2/n)) ) ⇒v_n =nlog(2n^(−(1/n)) {1−(1/2) n^(−(1/n)) }) =nlog(2 n^(−(1/n)) )+n log(1−(1/(2n^(1/n) ))) =nlog2−logn +nlog(1−(1/(2n^(1/n) ))) 2n^(1/n) =2 e^((logn)/n) →2 ⇒log(1−(1/(2n^(1/n) )))→−nlog2 ⇒lim_(n→+∞) v_n =−∞ ⇒ lim_(n→+∞) U_n =e^(−∞) =0
$$\mathrm{U}_{\mathrm{n}} =\frac{\left(\mathrm{2}\:\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:\Rightarrow\:\mathrm{U}_{\mathrm{n}} =\frac{\left(\mathrm{2n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}^{\frac{\mathrm{2}}{\mathrm{n}}} \right)^{\mathrm{n}} }\:=\left(\frac{\mathrm{2n}^{\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{1}}{\mathrm{n}^{\frac{\mathrm{2}}{\mathrm{n}}} }\right)^{\mathrm{n}} \\ $$$$=\left(\mathrm{2}\:\mathrm{n}^{−\frac{\mathrm{1}}{\mathrm{n}}} \:−\mathrm{n}^{−\frac{\mathrm{2}}{\mathrm{n}}} \right)^{\mathrm{n}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\mathrm{nlog}\left(\mathrm{2}\:\mathrm{n}^{−\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{n}^{\frac{−\mathrm{2}}{\mathrm{n}}} \right)} \\ $$$$\mathrm{v}_{\mathrm{n}} =\mathrm{nlog}\left(\mathrm{2}.\mathrm{n}^{−\frac{\mathrm{1}}{\mathrm{n}}} −\mathrm{n}^{−\frac{\mathrm{2}}{\mathrm{n}}} \right)\:\Rightarrow\mathrm{v}_{\mathrm{n}} =\mathrm{nlog}\left(\mathrm{2n}^{−\frac{\mathrm{1}}{\mathrm{n}}} \left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{n}^{−\frac{\mathrm{1}}{\mathrm{n}}} \right\}\right) \\ $$$$=\mathrm{nlog}\left(\mathrm{2}\:\mathrm{n}^{−\frac{\mathrm{1}}{\mathrm{n}}} \right)+\mathrm{n}\:\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2n}^{\frac{\mathrm{1}}{\mathrm{n}}} }\right)\:=\mathrm{nlog2}−\mathrm{logn}\:+\mathrm{nlog}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2n}^{\frac{\mathrm{1}}{\mathrm{n}}} }\right) \\ $$$$\mathrm{2n}^{\frac{\mathrm{1}}{\mathrm{n}}} \:=\mathrm{2}\:\mathrm{e}^{\frac{\mathrm{logn}}{\mathrm{n}}} \:\rightarrow\mathrm{2}\:\Rightarrow\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2n}^{\frac{\mathrm{1}}{\mathrm{n}}} }\right)\rightarrow−\mathrm{nlog2}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{v}_{\mathrm{n}} =−\infty\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{−\infty} \:=\mathrm{0} \\ $$

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