Science:Math Exam Resources/Courses/MATH105/April 2016/Question 04 (b)/Solution 1

We can rewrite the function $f(x)$ in the power series form as follows:

$f(x)=\sum _{k=1}^{\infty }a_{k}$, where $a_{k}={\frac {x^{k}}{k3^{k}}}.$

In order to find the radius of convergence for a power series, we first apply the ratio test. So,

$L=\lim \limits _{k\to \infty }{\Big |}{\frac {a_{k+1}}{a_{k}}}{\Big |}=\lim \limits _{k\to \infty }{\Big |}{\frac {\frac {x^{k+1}}{(k+1)3^{k+1}}}{\frac {x^{k}}{k3^{k}}}}{\Big |}=\lim \limits _{k\to \infty }{\Big |}{\frac {k}{k+1}}\cdot {\frac {3^{k}}{3^{k+1}}}\cdot x{\Big |}={\frac {1}{3}}|x|\lim \limits _{k\to \infty }{\Big |}{\frac {k}{k+1}}{\Big |}={\frac {1}{3}}|x|\cdot 1={\frac {1}{3}}|x|.$

Then, the power series converges when ${\frac {1}{3}}|x|<1$, (i.e., $|x|<3$), while it diverges when ${\frac {1}{3}}|x|>1$, (i.e., $|x|>3$). Therefore, the radius of convergence is ${\color {blue}R=3}.$

Now, we consider the second part. From the first part, we know that the interval of convergence of $f$ is $(-3,3)$, and $1$ is in this interval, also $f$ is differentiable at $1$ and its derivative is the derivative of the power series $f(x)$ term by term;

$g(x)=f'(x)={\frac {df}{dx}}={\frac {d}{dx}}\sum _{k=1}^{\infty }{\frac {1}{k3^{k}}}x^{k}=\sum _{k=1}^{\infty }{\frac {d}{dx}}{\frac {1}{k3^{k}}}x^{k}=\sum _{k=1}^{\infty }{\frac {1}{3^{k}}}x^{k-1}.$

We now compute $g(1):$

${\color {blue}g(1)}=\sum _{k=1}^{\infty }{\frac {1}{3^{k}}}(1)^{k-1}=\sum _{k=1}^{\infty }{\frac {1}{3^{k}}}={\frac {1}{3}}+{\frac {1}{3^{2}}}+{\frac {1}{3^{3}}}+{\frac {1}{3^{4}}}+...={\frac {\frac {1}{3}}{1-{\frac {1}{3}}}}={\frac {\frac {1}{3}}{\frac {2}{3}}}{\color {blue}={\frac {1}{2}}}.$

Note that the last series is a geometric series with the first term ${\frac {1}{3}}$ and the common ratio ${\frac {1}{3}}.$

To sum up, the answers are $\color {blue}{R=3,g(1)={\frac {1}{2}}}$.