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Question Number 194851 by Guriya last updated on 17/Jul/23

$$\mathbf{Name}\mathbf{Zainab}\mathbf{Bibi}$$$$\mathbf{BC}200400692$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathbf{Mth}621$$$$\mathrm{solution}..$$$${\mathrm{a}}_{\mathrm{n}}=\frac{(\mathrm{n}-)!}{{(\mathrm{n}+)}^2}$$$${\mathrm{a}}_{\mathrm{n}+}=\frac{(\mathrm{n}+-)!}{{(\mathrm{n}++)}^2}=\frac{\left(\mathrm{n}\right)!}{{(\mathrm{n}+2)}^2}$$$$\mathrm{by}\mathrm{ratio}\mathrm{test}$$$$\frac{{\mathrm{a}}_{\mathrm{n}+}}{{\mathrm{a}}_{\mathrm{n}}}=\frac{\frac{\left(\mathrm{n}\right)!}{(\mathrm{n}+2)}}{\frac{(\mathrm{n}-)!}{{(\mathrm{n}+)}^2}}=\frac{\left(\mathrm{n}\right)!}{{(\mathrm{n}+2)}^2}\times \frac{{(\mathrm{n}+)}^2}{(\mathrm{n}+)!}$$$$\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{{\mathrm{a}}_{\mathrm{n}+}}{{\mathrm{a}}_{\mathrm{n}}}=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{\mathrm{n}(\mathrm{n}-)!}{{(\mathrm{n}+2)}^2}\times \frac{{(\mathrm{n}+)}^2}{(\mathrm{n}-)!}\because \left(\mathrm{n}\right)!=\mathrm{n}(\mathrm{n}-)!$$$$\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{{\mathrm{a}}_{\mathrm{n}+}}{{\mathrm{a}}_{\mathrm{n}}}=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{\mathrm{n}{(\mathrm{n}+)}^2}{{(\mathrm{n}+2)}^2}=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{\mathrm{n}({\mathrm{n}}^2+1+2\mathrm{n})}{{\mathrm{n}}^2+4+4\mathrm{n}}$$$$\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{({\mathrm{n}}^3+\mathrm{n}+{2\mathrm{n}}^2)}{{\mathrm{n}}^2+4+4\mathrm{n}}=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{(1+\frac{1}{{\mathrm{n}}^2}+\frac{2}{\mathrm{n}})}{\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^3}+\frac{4}{{\mathrm{n}}^2}}$$$$\mathrm{Divided}\mathrm{by}{\mathrm{n}}^3\mathrm{to}\mathrm{numerator}\mathrm{and}\mathrm{denominator}$$$$\mathrm{Now}\mathrm{by}\mathrm{Applying}\mathrm{limit}$$$$\frac{(1+\frac{1}{{\mathrm{\infty }}^2}+\frac{2}{\mathrm{\infty }})}{\frac{1}{\mathrm{\infty }}+\frac{4}{{\mathrm{\infty }}^3}+\frac{4}{{\mathrm{\infty }}^2}}=\frac{1+0+0}{0+0+0}=\frac{1}{0}=\mathrm{\infty }$$$$\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}=\mathrm{\infty }$$

Commented by Frix last updated on 17/Jul/23

$$\frac{1}{0}=\mathrm{\infty }\mathrm{is}\mathrm{not}\mathrm{allowed};\mathrm{instead}\mathrm{use}\mathrm{this}:$$$$\frac{{\mathrm{a}}_{\mathrm{n}+}}{{\mathrm{a}}_{\mathrm{n}}}=\frac{\frac{\left(\mathrm{n}\right)!}{(\mathrm{n}+2)}}{\frac{(\mathrm{n}-)!}{{(\mathrm{n}+)}^2}}=\frac{\left(\mathrm{n}\right)!}{{(\mathrm{n}+2)}^2}\times \frac{{(\mathrm{n}+)}^2}{(\mathrm{n}-)!}=$$$$=n\times \frac{n^2+2n+1}{n^2+4n+4}=n\times (1+\frac{2n+1}{n^2+4n+1})$$$$\underset{n\to \mathrm{\infty }}{\lim }n=\mathrm{\infty }$$$$\underset{n\to \mathrm{\infty }}{\lim }1=1$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{2n+1}{n^2+4n+1}=0$$$$\mathrm{\infty }\times (1+0)=\mathrm{\infty }$$

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