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Question Number 203774 by Calculusboy last updated on 27/Jan/24

determine whether the series is convergent or divergent 𝚺_(n=1) ^∞ (n/( (√(4n^2 +1))))

$$\boldsymbol{{determine}}\:\boldsymbol{{whether}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{convergent}}\:\boldsymbol{{or}}\:\boldsymbol{{divergent}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{{n}}}{\:\sqrt{\mathrm{4}\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}} \\ $$

Answered by witcher3 last updated on 27/Jan/24

(n/( (√(4n^2 +1))))β‰₯(n/( (√(1+4n^2 +4n))))=(n/(2n+1))=((3n)/(3(2n+1)))>((2n+1)/(3(2n+1))) β‡’Ξ£(n/( (√(4n^2 +1))))>Ξ£(1/3) Dv

$$\frac{\mathrm{n}}{\:\sqrt{\mathrm{4n}^{\mathrm{2}} +\mathrm{1}}}\geqslant\frac{\mathrm{n}}{\:\sqrt{\mathrm{1}+\mathrm{4n}^{\mathrm{2}} +\mathrm{4n}}}=\frac{\mathrm{n}}{\mathrm{2n}+\mathrm{1}}=\frac{\mathrm{3n}}{\mathrm{3}\left(\mathrm{2n}+\mathrm{1}\right)}>\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{3}\left(\mathrm{2n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{n}}{\:\sqrt{\mathrm{4n}^{\mathrm{2}} +\mathrm{1}}}>\Sigma\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{Dv} \\ $$$$ \\ $$

Answered by Mathspace last updated on 30/Jan/24

(n/( (√(4n^2 +1))))=(n/(2n(√(1+(1/(4n^2 ))))))=(1/(2(√(1+(1/(4n^2 )))))) we have (1+x)^Ξ± =1+Ξ±x +((Ξ±(Ξ±βˆ’1))/2)x^2 +o(x^2 )β‡’ (1/( (√(1+x))))=(1+x)^(βˆ’(1/2)) =1βˆ’(x/2)+(1/2)(βˆ’(1/2))(βˆ’(1/2)βˆ’1)x^2 +o(x^2 ) =1βˆ’(x/2)βˆ’(1/4)Γ—((βˆ’3)/2) x^2 +o(x^2 ) =1βˆ’(x/2)+(3/8)x^2 +o(x^2 )so u_n =(1/( 2(√(1+(1/(4n^2 ))))))=(1/2)βˆ’(1/(16n^2 ))+(3/(16))Γ—(1/(16n^4 ))+o((1/n^4 )) est un dev.assymtotique de u_n ona lim u_n =(1/2) β‰ 0 β‡’ Ξ£u_n est divergente.

$$\frac{{n}}{\:\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}}=\frac{{n}}{\mathrm{2}{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }}} \\ $$$${we}\:{have}\:\left(\mathrm{1}+{x}\right)^{\alpha} =\mathrm{1}+\alpha{x}\:+\frac{\alpha\left(\alphaβˆ’\mathrm{1}\right)}{\mathrm{2}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)\Rightarrow \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}=\left(\mathrm{1}+{x}\right)^{βˆ’\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{1}βˆ’\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left(βˆ’\frac{\mathrm{1}}{\mathrm{2}}\right)\left(βˆ’\frac{\mathrm{1}}{\mathrm{2}}βˆ’\mathrm{1}\right){x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}βˆ’\frac{{x}}{\mathrm{2}}βˆ’\frac{\mathrm{1}}{\mathrm{4}}Γ—\frac{βˆ’\mathrm{3}}{\mathrm{2}}\:{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}βˆ’\frac{{x}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{8}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right){so} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\mathrm{2}}βˆ’\frac{\mathrm{1}}{\mathrm{16}{n}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{16}}Γ—\frac{\mathrm{1}}{\mathrm{16}{n}^{\mathrm{4}} }+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\right) \\ $$$${est}\:{un}\:{dev}.{assymtotique}\:{de} \\ $$$${u}_{{n}} \:\:\:\:{ona}\:{lim}\:{u}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\neq\mathrm{0}\:\Rightarrow \\ $$$$\Sigma{u}_{{n}} \:{est}\:{divergente}. \\ $$$$ \\ $$

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