Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence. 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?

Write $x=1+t$, so you need to find the Taylor series of
$$
\frac{1}{2}(1+t)^3\ln(1+t)
$$
at $t=0$. Since
$$
\ln(1+t)=\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}
$$
your Taylor series is
$$
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+
\frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+1}}{n}+
\frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+2}}{n}+
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+3}}{n}
$$
that can be rewritten as
$$
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+
\frac{3}{2}\sum_{n>1}\frac{(-1)^{n}t^{n}}{n-1}+
\frac{3}{2}\sum_{n>2}\frac{(-1)^{n-1}t^{n}}{n-2}+
\frac{1}{2}\sum_{n>3}\frac{(-1)^{n-2}t^{n}}{n-3}
$$
Now isolate the only term for $n=1$, the two terms for $n=2$ and the three terms for $n=3$:
$$
\frac{t}{2}+\frac{1}{2}\left(-\frac{t^2}{2}+3t^2\right)+
\frac{1}{2}\left(\frac{t^3}{3}-\frac{3t^3}{2}+3t^3\right)+\\
\frac{1}{2}\sum_{n>3}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n3}{n-1}+\frac{(-1)^{n-1}3}{n-2}+\frac{(-1)^{n-2}}{n-3}\right)t^n
$$
The last summation can be rewritten as
$$
\frac{1}{2}\sum_{n>3}\left(-\frac{1}{n}+\frac{3}{n-1}-\frac{3}{n-2}+\frac{1}{n-3}\right)(-1)^nt^n=
\sum_{n>3}\frac{3(-1)^nt^n}{n(n-1)(n-2)(n-3)}
$$
so the end result is
$$
\frac{t}{2}+\frac{5}{4}t^2+\frac{11}{12}t^3+\sum_{n>3}\frac{3(-1)^nt^n}{n(n-1)(n-2)(n-3)}
$$
and you can check that the coefficient for $t^4$ is indeed $1/8$; the coefficient for $t^5$ is $-1/40$ as you computed.
The Taylor series for $f(x)=x^3\ln\sqrt{x}$ around $1$ is obtained by substituting $t$ with $x-1$. The radius of convergence is the same as for $\ln(1+t)$, that is, $1$. You can also compute it with the ratio test, as further check.
