TITLE: Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence.
QUESTION [7 upvotes]: As the question states: 
"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."
I have consulted this related question, and understand the steps to be:

find the first few terms of the Taylor polynomial.
Generalize the terms by making use of an infinite sum to represent the function as the Taylor series.
use the infinite sum in the ratio test to find the radius of convergence.


Progress so far:

The first 6 terms (n = 0 to n = 5) of the Taylor polynomial I have calculated to be:

$x^3 \cdot \ln{(\sqrt{x})} + \frac{1}{2}(x-a) + \frac{5}{4}(x-a)^2 + \frac{11}{12}(x-a)^3 + \frac{1}{8}(x-a)^4 - \frac{1}{40}(x-a)^5$
It is at this point however that I fall over. It is not intuitive to me how I can write the c-terms as a function without utilizing some sort of online maths engine for fitting the data to a curve.
Is there some sort of first-year-student-friendly technique for modelling these data points systematically? Alternatively, does someone have an intuition they would be willing to share for solving this problem?

REPLY [3 votes]: In fact a lot of the work needed is already done. Let's recapitulate. We have $f(x)=x^3\ln\left(\sqrt{ x}\right)$ and the Taylor expansion at $a=1$ is given as
\begin{align*}
f(x)=\sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(a)(x-a)^n
\end{align*}

We obtain
  \begin{align*}
f(x)&=x^3\ln\left(\sqrt{ x}\right)=\frac{1}{2}x^3\ln(x)\\
f^{\prime}(x)&=\frac{3}{2}x^2\ln(x)+\frac{1}{2}x^2\\
f^{\prime\prime}(x)&=3x\ln(x)+\frac{5}{2}x\\
f^{\prime\prime\prime}(x)&=3\ln(x)+\frac{11}{2}\\
f^{(4)}(x)&=\frac{3}{x}\\
f^{(5)}(x)&=-\frac{3}{x^2},\qquad f^{(6)}(x)=3\cdot\frac{2!}{x^3},\qquad f^{(7)}=-3\cdot \frac{3!}{x^4}\\
&\vdots\\
f^{(n)}(x)&=3(-1)^n\frac{(n-4)!}{x^{n-3}}\qquad\qquad n\geq 4\tag{1}
\end{align*}

From the fourth derivative $\frac{3}{x}$ we can relatively easy obtain higher derivatives and assume the general formula (1) which can be shown by induction. 

Evaluated at $a=1$ we have
  \begin{align*}
f(1)=0, f^{\prime}(1)=\frac{1}{2}, f^{\prime\prime}(1)=\frac{5}{2}, f^{\prime\prime\prime}(1)=\frac{11}{2}, f^{(n)}(1)=3(-1)^n(n-4)!\qquad n\geq 4
\end{align*}
We obtain from the derivatives above the Taylor series
  \begin{align*}
\color{blue}{f(x)}&\color{blue}{=x^3\ln\sqrt{ x}}\\
&\color{blue}{=\frac{1}{2}(x-1)+\frac{5}{4}(x-1)^2+\frac{11}{12}(x-1)^3+3\sum_{n=4}^\infty\frac{(-1)^n}{n(n-1)(n-2)(n-3)}(x-1)^n}
\end{align*}
The radius $R$ of convergence is
  \begin{align*}
R&=\lim_{n\to \infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim_{n\to \infty}\left|\frac{3(-1)^n(n+1)n(n-1)(n-2)}{3(-1)^{n+1}n(n-1)(n-2)(n-3)}\right|\\
&=\lim_{n\to\infty}\left|\frac{n+1}{n-3}\right|\\
&=1
\end{align*}