Question Number 205367 by hardmath last updated on 18/Mar/24
Answered by Ghisom last updated on 18/Mar/24
$$\left(\frac{{n}^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}^{{n}} }+\frac{{n}^{\mathrm{2}} }{\mathrm{4}^{{n}} }+\frac{{n}^{\mathrm{5}/\mathrm{2}} }{\mathrm{5}^{{n}} }\right)^{\mathrm{1}/{n}} = \\ $$$$=\frac{{n}^{\mathrm{3}/\left(\mathrm{2}{n}\right)} }{\mathrm{60}}\left(\mathrm{20}^{{n}} +\mathrm{15}^{{n}} {n}^{\mathrm{1}/\mathrm{2}} +\mathrm{12}^{{n}} {n}\right)^{\mathrm{1}/{n}} \\ $$$${n}\rightarrow\infty\:\Rightarrow\:\begin{cases}{{n}^{\mathrm{3}/\left(\mathrm{2}{n}\right)} \rightarrow\mathrm{1}}\\{\left(\mathrm{20}^{{n}} +…\right)^{\mathrm{1}/{n}} \rightarrow\mathrm{20}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{lim}\:=\:\frac{\mathrm{20}}{\mathrm{60}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by hardmath last updated on 21/Mar/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{sir} \\ $$
Commented by Ghisom last updated on 22/Mar/24
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$