Can you?part:5

Calculus Level 3

Compute $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { { (n!) }^{ e } }^{( n) } } }$ till 2 decimal places.

It is n!^e^n in denominator.

The answer is 1.00.

1 solution

Alex Li
Feb 19, 2015

Note that all terms with $n>2$ are negligibly small at the required level of precision, because $(3!)^{e^3}>6^{20}>10^{15}$ . The sum of the first 2 terms is $1+\frac{1}{2^{e^2}}\approx1.005966$ , which rounds to $\boxed{1.01}$ .

Can you?part:5

The answer is 1.00.

1 solution

0 pending reports