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2 years ago

6

Solve for k. K(K + 8) = 0 Write your answers as integers or as proper or improper fractions in simplest form. k = 1 or k = submi

t

1 answer:

Alenkasestr [34]2 years ago

3 0

Answer:

k= -8

Step-by-step explanation:

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Step-by-step explanation:

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1 year ago

<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%

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We have

$\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)$

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

$\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}$

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1 year ago