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Question Number 97418 by Rio Michael last updated on 08/Jun/20

Verify if the series Σ_(n=1) ^n ((2n + 5)/(n^2 +3n + 2)) is convergent or divergent. What method is easier?

$$\mathrm{Verify}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\: \\ $$$$\:\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}{n}\:+\:\mathrm{5}}{{n}^{\mathrm{2}} \:+\mathrm{3}{n}\:+\:\mathrm{2}}\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{or}\:\mathrm{divergent}. \\ $$$$\mathrm{What}\:\mathrm{method}\:\mathrm{is}\:\mathrm{easier}? \\ $$

Commented by Tony Lin last updated on 08/Jun/20

Σ_(n=1) ^∞ (((n+1)+(n+2)+2)/((n+1)(n+2))) =Σ_(n=1) ^∞ (1/(n+2))+Σ_(n=1) ^∞ (1/(n+1))+2Σ_(n=1) ^∞ (1/((n+1)(n+2))) =2Σ_(n=1) ^∞ (1/n)−(1/2)−1−(1/2)+1 =2Σ_(n=1) ^∞ (1/n)−1 Σ_(n=1) ^∞ (1/n) is divergent ⇒Σ_(n=1) ^∞ ((2n+5)/(n^2 +3n+2)) is divergent

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+\mathrm{1}\right)+\left({n}+\mathrm{2}\right)+\mathrm{2}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\: \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{2}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\mathrm{1} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\:{is}\:{divergent} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{5}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\:{is}\:{divergent} \\ $$

Answered by mathmax by abdo last updated on 08/Jun/20

u_n =((2n+5)/(n^2 +3n+2)) ⇒u_n =((n(2+(5/n)))/(n^2 (1+(3/n)+(2/n^2 )))) ⇒u_n ∼((2+(5/n))/(n(1+(3/n)+(2/n^2 )))) ∼(2/n) Σ (2/n) diverges ⇒Σ u_n diverges

$$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{2n}+\mathrm{5}}{\mathrm{n}^{\mathrm{2}} \:+\mathrm{3n}+\mathrm{2}}\:\Rightarrow\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{n}\left(\mathrm{2}+\frac{\mathrm{5}}{\mathrm{n}}\right)}{\mathrm{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{2}}{\mathrm{n}^{\mathrm{2}} }\right)}\:\Rightarrow\mathrm{u}_{\mathrm{n}} \:\sim\frac{\mathrm{2}+\frac{\mathrm{5}}{\mathrm{n}}}{\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{n}}+\frac{\mathrm{2}}{\mathrm{n}^{\mathrm{2}} }\right)}\:\sim\frac{\mathrm{2}}{\mathrm{n}} \\ $$$$\Sigma\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{diverges}\:\Rightarrow\Sigma\:\mathrm{u}_{\mathrm{n}} \:\mathrm{diverges} \\ $$

Commented by Rio Michael last updated on 08/Jun/20

thank you sir,but sir do you have any other method?

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir},\mathrm{but}\:\mathrm{sir}\:\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{any}\:\mathrm{other}\:\mathrm{method}? \\ $$

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