Evaluate $\underset{n\to \mathrm{\infty}}{lim}\frac{{e}^{{n}^{2}}}{(2n)!}$

daktielti
2022-06-29
Answered

Evaluate $\underset{n\to \mathrm{\infty}}{lim}\frac{{e}^{{n}^{2}}}{(2n)!}$

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Alexzander Bowman

Answered 2022-06-30
Author has **19** answers

perhaps it becomes clearer that the limit is infinite, making the series expansion of the exponential

$\frac{{e}^{{n}^{2}}}{(2n)!}}={\displaystyle \frac{\sum _{m=0}^{\mathrm{\infty}}{\displaystyle \frac{({n}^{2}{)}^{m}}{m!}}}{(2n)!}}\ge {\displaystyle \frac{{n}^{4n}}{((2n)!{)}^{2}}}={\left({\displaystyle \frac{{n}^{2n}}{(2n)!}}\right)}^{2$

the last expression on the right is $\ge Cn$ $C>0$ for large n.

$\frac{{e}^{{n}^{2}}}{(2n)!}}={\displaystyle \frac{\sum _{m=0}^{\mathrm{\infty}}{\displaystyle \frac{({n}^{2}{)}^{m}}{m!}}}{(2n)!}}\ge {\displaystyle \frac{{n}^{4n}}{((2n)!{)}^{2}}}={\left({\displaystyle \frac{{n}^{2n}}{(2n)!}}\right)}^{2$

the last expression on the right is $\ge Cn$ $C>0$ for large n.

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