The function $$\frac {\cos \pi x}{x}$$ is decreasing for $x \geq \answer {N/A}$ (if the function is not ultimately decreasing, enter N/A). The integral test apply to the series $$\sum _{n=1}^\infty \frac {\cos \pi n}{n}.$$ The series .

The function $$\frac {\ln x}{x}$$ is decreasing for $x \geq \answer {e}$ (if the function is not ultimately decreasing, enter N/A). The integral test apply to the series $$\sum _{n=3}^\infty \frac {\ln n}{n}.$$ We have $$\int _3^\infty \frac {\ln x}{x} dx = \answer {\infty },$$ so the series .

The function $$\frac {1}{x^2+1}$$ is decreasing for $x \geq \answer {0}$ (if the function is not ultimately decreasing, enter N/A). The integral test apply to the series $$\sum _{n=1}^\infty \frac {1}{n^2+1}.$$ We have that $$\int _1^\infty \frac {dx}{x^2 + 1} = \answer {\frac {\pi }{4}},$$ so the series .

  Reveal Hint (1 of 1) ( 1) The integral test apply to the series $$\sum _{n=2}^\infty \frac {1}{n (\ln n)^2}.$$ The series .

  Reveal Hint (1 of 1) ( 1) The integral test apply to the series $$\sum _{n=2}^\infty \frac {1}{n (\ln n)}.$$ The series .

The integral test apply to the series $$\sum _{n=2}^\infty e^{-n^2} \cos n \pi .$$ The series .

  Reveal Hint (1 of 1) ( 1) The integral test apply to the series $$\sum _{n=2}^\infty n^3 e^{-n^2}.$$ The series .

  Reveal Hint (1 of 1) ( 1) The function $$\frac {x}{(x^2+3)^2}$$ is decreasing for $x \geq \answer {1}$. By the integral test, $$\int _{\answer {8}}^\infty \frac {x}{(x^2+3)^2} dx \leq \sum _{n=8}^\infty \frac {n}{(n^2+3)^2} \leq \int _{\answer {7}}^\infty \frac {x}{(x^2+3)^2} dx.$$ We can approximate the infinite series by the sum of the first seven terms with what bounds on the error? $$\frac {1}{\answer {134}} + \sum _{n=1}^7 \frac {n}{(n^2+3)^2} \leq \sum _{n=1}^\infty \frac {n}{(n^2+3)^2} \leq \frac {1}{\answer {104}} + \sum _{n=1}^7 \frac {n}{(n^2+3)^2}.$$

  Reveal Hint (1 of 1) ( 1) Using the integral test, we can determine that the sum of the series $$\sum _{n=1}^\infty \frac {1}{n^2}$$ is $2$.