The Maclaurin series for a function $f$ is given by $\sum_{n=1}^{\infty} \frac{(n+1)x^n}{n^2 6^n}$ and converges to $f(x)$ for all $x$ in the interval of convergence. It can be shown that the Maclaurin series for $f$ has a radius of convergence of 6. (a) Determine whether the Maclaurin series for $f$ converges or diverges at $x = 6$. Give a reason for your answer. (b) It can be shown that $f(-3) = \sum_{n=1}^{\infty} \frac{(n+1)(-3)^n}{n^2 6^n} = \sum_{n=1}^{\infty} \frac{n+1}{n^2}\left(-\frac{1}{2}\right)^n$ and that the first three terms of this series sum to $S_3 = -\frac{125}{144}$. Show that $\left|f(-3) - S_3\right| < \frac{1}{50}$. (c) Find the general term of the Maclaurin series for $f'$, the derivative of $f$. Find the radius of convergence of the Maclaurin series for $f'$. (d) Let $g(x) = \sum_{n=1}^{\infty} \frac{(n+1)x^{2n}}{n^2 3^n}$. Use the ratio test to determine the radius of convergence of the Maclaurin series for $g$.
The Maclaurin series for a function $f$ is given by $\sum_{n=1}^{\infty} \frac{(n+1)x^n}{n^2 6^n}$ and converges to $f(x)$ for all $x$ in the interval of convergence. It can be shown that the Maclaurin series for $f$ has a radius of convergence of 6.
(a) Determine whether the Maclaurin series for $f$ converges or diverges at $x = 6$. Give a reason for your answer.
(b) It can be shown that $f(-3) = \sum_{n=1}^{\infty} \frac{(n+1)(-3)^n}{n^2 6^n} = \sum_{n=1}^{\infty} \frac{n+1}{n^2}\left(-\frac{1}{2}\right)^n$ and that the first three terms of this series sum to $S_3 = -\frac{125}{144}$. Show that $\left|f(-3) - S_3\right| < \frac{1}{50}$.
(c) Find the general term of the Maclaurin series for $f'$, the derivative of $f$. Find the radius of convergence of the Maclaurin series for $f'$.
(d) Let $g(x) = \sum_{n=1}^{\infty} \frac{(n+1)x^{2n}}{n^2 3^n}$. Use the ratio test to determine the radius of convergence of the Maclaurin series for $g$.