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

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

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

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{(n+1)+(n+2)+2}{(n+1)(n+2)}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}+2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)(n+2)}$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{2}-1-\frac{1}{2}+1$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-1$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}isdivergent$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n+5}{n^2+3n+2}isdivergent$$

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

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

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

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

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