Question Number 140982 by Mathspace last updated on 14/May/21
$${convergence}\:{and}\:{value}\:{of} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{{n}} }{\left({n}!\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$
Commented by mohammad17 last updated on 14/May/21
$${U}_{{n}} =\frac{{n}^{{n}} }{\left({n}!\right)^{\mathrm{2}} }\:\:\:\:\:,\:\:{U}_{{n}+\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${lim}_{{n}\rightarrow\infty} \frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }={lim}_{{n}\rightarrow\infty} \frac{\left({n}+\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}!\right)^{\mathrm{2}} }×\frac{\left({n}!\right)^{\mathrm{2}} }{{n}^{{n}} } \\ $$$$ \\ $$$$={lim}_{{n}\rightarrow\infty} \left(\frac{{n}+\mathrm{1}}{{n}}\right)^{{n}} ×{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}={e}^{{n}} ×\mathrm{0}=\mathrm{0} \\ $$$$ \\ $$$${So},{the}\:{series}\:{is}\:{converge}\:{by}\:{ratio}\:{test} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 14/May/21
$$\mathrm{why}\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \mathrm{e}^{\mathrm{n}} .\mathrm{0}=\mathrm{0}? \\ $$
Commented by mnjuly1970 last updated on 14/May/21
$$\:\:{e}.\mathrm{0}=\mathrm{0} \\ $$