Question: Find the sum of the series ∑n=1∞12n12 correct to three decimal places.Step 1Consider that f(x)=12x12 is positive and continuous for x>0.To decide if f(x)=12x12 is also decreasing, we can examine the derivative f'(x)=-144x13,-144x13.Step 2Examining the derivative, we have f'(x)=-144x-13=-144x13.Since the denominator is always positive on (0,∞) then

Find the sum $of$ the series ${\displaystyle \sum _{n=1}^{\infty }}\frac{12}{n^{12}}$ correct $to$ three decimal places.

Step $1$

Consider that $f(x)=\frac{12}{x^{12}}is$ positive and continuous for $x>0.$

$To$ decide $iff(x)=\frac{12}{x^{12}}is$ also decreasing, $we$ can examine the derivative $f^{\text{'}}(x)=-\frac{144}{x^{13}},-\frac{144}{x^{13}}.$

Step $2$

Examining the derivative, $we$ have $f^{\text{'}}(x)=-144x^{-13}=\frac{-144}{x^{13}}.$

Since the denominator $is$ always positive $on(0,\infty )$ then $\frac{-144}{x^{13}}is$ always negative negative.

Step $3$

Since $f^{\text{'}}(x)is$ always negative, then $f(x)=\frac{12}{x^{12}}is$ decreasing $on(0,\infty ).$