Maybe this is not Calculus

Calculus Level 3

$\large \lim_{n \to \infty } \frac{(n!)^{2}}{\displaystyle \sum_{k=1}^{n} \left(\frac{n!}{k} \right )^{2}}$

If the value of the limit above is $x$ , and if the expression $\displaystyle \frac { 4\left \lfloor \pi^{2}x \right \rfloor + 1 }{7 }$ can be expressed in the form $\large \frac {a^{2}}{b}$ where $a$ and $b$ are coprime positive integers. Determine $a + b$ .

The answer is 12.

1 solution

Jake Lai
May 21, 2015

$n!$ is a constant within the limit, so the numerator and denominator cancel out to give

$\lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n} \frac{1}{k^{2}} \right)^{-1} = \frac{6}{\pi^{2}}$

which everyone knows because it's 2015 guys. So,

$\frac{4\lfloor \pi^{2}x \rfloor +1}{7} = \frac{25}{7} = \frac{5^{2}}{7}$

Hence, $a+b = 5+7 = \boxed{12}$ .

Everyone knows because this is " brilliant.org " not beacuse this is " 2015 " .

Nishu sharma - 6 years ago

Well, true.

Jake Lai - 6 years ago

Maybe this is not Calculus

The answer is 12.

1 solution

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