The Maclaurin series for the function $f$ is given by $f ( x ) = \sum _ { k = 1 } ^ { \infty } \frac { ( - 1 ) ^ { k + 1 } x ^ { k } } { k ^ { 2 } } = x - \frac { x ^ { 2 } } { 4 } + \frac { x ^ { 3 } } { 9 } - \cdots$ on its interval of convergence. (a) Use the ratio test to determine the interval of convergence of the Maclaurin series for $f$. Show the work that leads to your answer. (b) The Maclaurin series for $f$ evaluated at $x = \frac { 1 } { 4 }$ is an alternating series whose terms decrease in absolute value to 0. The approximation for $f \left( \frac { 1 } { 4 } \right)$ using the first two nonzero terms of this series is $\frac { 15 } { 64 }$. Show that this approximation differs from $f \left( \frac { 1 } { 4 } \right)$ by less than $\frac { 1 } { 500 }$. (c) Let $h$ be the function defined by $h ( x ) = \int _ { 0 } ^ { x } f ( t ) \, d t$. Write the first three nonzero terms and the general term of the Maclaurin series for $h$.
The Maclaurin series for the function $f$ is given by $f ( x ) = \sum _ { k = 1 } ^ { \infty } \frac { ( - 1 ) ^ { k + 1 } x ^ { k } } { k ^ { 2 } } = x - \frac { x ^ { 2 } } { 4 } + \frac { x ^ { 3 } } { 9 } - \cdots$ on its interval of convergence.\\
(a) Use the ratio test to determine the interval of convergence of the Maclaurin series for $f$. Show the work that leads to your answer.\\
(b) The Maclaurin series for $f$ evaluated at $x = \frac { 1 } { 4 }$ is an alternating series whose terms decrease in absolute value to 0. The approximation for $f \left( \frac { 1 } { 4 } \right)$ using the first two nonzero terms of this series is $\frac { 15 } { 64 }$. Show that this approximation differs from $f \left( \frac { 1 } { 4 } \right)$ by less than $\frac { 1 } { 500 }$.\\
(c) Let $h$ be the function defined by $h ( x ) = \int _ { 0 } ^ { x } f ( t ) \, d t$. Write the first three nonzero terms and the general term of the Maclaurin series for $h$.