MacLaurin Expansion of $\log(1+x) \cdot e^x$ Could anyone enlighten me on how to go about expanding the following function around $x_0 = 0$:
$$
f(x):= \log(1+x)e^{x}
$$
I have tried using Cauchy Product Series and bruteforce computation of the coefficients but I always find myself with a double sum which is difficult to work with when trying to estimate the radius of convergence.

As always any comment or answer is welcome and let me know if I can explain myself clearer!
 A: We are looking of the McLaurin series
$$
f(x)=\log(1+x)e^x \quad\Longrightarrow\quad f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n
$$
of $f(x)=\log(1+x)e^x$.
So
$$
f^{(n)}(x)=\sum_{k=0}^n 
\binom{n}{k}\big(\log(1+x)\big)^{(k)}(e^x)^{(n-k)}=\log(1+x)e^x+\sum_{k=1}^n 
\binom{n}{k}\frac{(-1)^{k-1}(k-1)!}{(1+x)^k}e^x
$$
Hence
$$
f^{(n)}(0)=\sum_{k=1}^n 
\binom{n}{k}(-1)^{k-1}(k-1)!=\sum_{k=1}^n 
\frac{n!}{k(n-k)!}(-1)^{k-1}
$$
and thus
$$
\frac{f^{(n)}(0)}{n!}=\sum_{k=1}^n 
\frac{(-1)^{k-1}}{k(n-k)!}
$$
A: Note that $$
\frac{d^n}{dx^n} \log(1+x) = (-1)^{n-1}\frac{(n-1)!}{(1+x)^n}
$$
thus $$
f^{(n)}(x) = \log(1+x)e^x + \sum_{k=1}^n \binom{n}{k}(-1)^{k-1} \frac{(k-1)!}{(1+x)^n} e^x
$$
Evaluating at $0$ gives the $n$th term of the Taylor series as \begin{eqnarray}
a_n &=& \frac{f^{(n)}(0)}{n!} = \sum_{k=1}^n \binom{n}{k}(-1)^{k-1} \frac{(k-1)!}{n!}\\& =& \sum_{k=1}^n (-1)^{k-1} \frac{n!}{k!(n-k)!}\frac{(k-1)!}{n!} = \sum_{k=1}^n (-1)^{k-1} \frac{n!}{k!(n-k)!}\frac{(k-1)!}{n!}\\
&=&\sum_{k=1}^n(-1)^{k-1}\frac{1}{k(n-k)!}
\end{eqnarray}
I do not believe this will simplify to a nicer closed form.
As far as radius of convergence is concerned, note that $f$ blows up as $x$ approaches $-1$, and that $f$ is analytic for all complex $x$ with $|x|<1$. Therefore the radius of convergence of the Taylor series is $1$.
If that is a little too much complex analysis, you can also show the radius of convergence is $1$ directly:$$
|a_n| \le \sum_{k=1}^n \frac1{k(n-k)!} < \sum_{k=1}^n 1 = n
$$
Therefore $$
\text{Radius of convergence} = \frac{1}{\limsup\limits_{n\rightarrow\infty} |a_n|^{1/n}} \ge \frac{1}{\limsup\limits_{n\rightarrow\infty} n^{1/n} }= 1
$$
Because $\lim\limits_{x\rightarrow 1^+} f(x) = -\infty$, the series cannot converge to $f$ at $x=-1$, therefore the radius of convergence cannot be bigger than $1$, so it is $1$.

If you want a better handle on the asymptotics of $|a_n|$, you can show that $$
\lim\limits_{n\rightarrow\infty} n(-1)^n a_n = \frac1e
$$
from which getting $\limsup\limits_{n\rightarrow\infty} |a_n|^{1/n}=1$ is easy. To prove \begin{eqnarray}
(-1)^n n a_n &=& \sum_{k=1}^n (-1)^{n+k-1}\frac{n}{k(n-k)!} \\&=& \sum_{k=1}^n (-1)^{n-k-1}\frac{1}{(n-k)!}(\frac{n-k}k + 1)\\ &=& \sum_{k=1}^n (-1)^{n-k-1}\frac{1}{(n-k)!} + \sum_{k=1}^n (-1)^{n-k-1}\frac{1}{(n-k)!}\frac{n-k}k\\
&=& \sum_{j=0}^{n-1}\frac{(-1)^j}{j!} + \sum_{j=1}^{n-1} \frac{(-1)^j}{(j-1)!(n-j)}
\end{eqnarray}
The first sum converges to $1/e$ and the second goes to $0$ by DCT (bound each term by $1/j!$)
A: $$
\log(1+x) = \sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} x^k
\\
e^x = \sum_{j=0}^\infty\frac{1}{j!}x^j
\\
\log(1+x)e^x = \sum_{n=1}^\infty\left(\sum_{k+j=n}\frac{(-1)^{k+1}}{k\,j!}\right)x^n
=\sum_{n=1}^\infty\left(\sum_{k=1}^n\frac{(-1)^{k+1}}{k\,(n-k)!}\right)x^n
$$
So our Laurent series is $\sum_{n=1}^\infty c_n x^n$ where
$$
c_n = \sum_{k=1}^n\frac{(-1)^{k+1}}{k\,(n-k)!}
=\frac{{}_3F_1\big(1,1,1-n;2;1\big)}{(n-1)!},\qquad n = 1,2,3,\dots
$$
The first few terms:
$$
x+{\frac{1}{2}}{x}^{2}+{\frac{1}{3}}{x}^{3}
+0 x^4 +{\frac{3}{40}}{x}^{5}-{
\frac{7}{144}}{x}^{6}+{\frac{23}{504}}{x}^{7}-{\frac{29}{720}}{x}^{8}+
{\frac{629}{17280}}{x}^{9}+O \left( {x}^{10} \right)
$$
A: The product rule solution shown works very well (and thank you to the other answerers, you've done some lovely work) but for the sake of completeness I thought it would be good to show that your original idea of using the Cauchy product formula does work out, and gives the same result.
We can start with our bog-standard Maclaurin series for $e^x$ and $\log(1+x)$:
$$e^x = \sum_{k = 0}^\infty \frac{x^k}{k!} \\ \log(1+x)=\log(1-(-x))= -\sum_{k=0}^\infty\frac{(-x)^{k+1}}{k+1} = \sum_{k=0}^\infty \frac{(-1)^kx^{k+1}}{k+1}$$
So, using our Cauchy product formula, we know that the product can be written as
$$e^x\log(1+x) = \sum_{k = 0}^\infty \left(\sum_{l = 0}^k \frac{x^{l + 1}}{l+1} \cdot \frac{x^{k - l}}{(k - l)!}\right) = \sum_{k = 0}^\infty \sum_{l = 0}^k \frac{(-1)^kx^{k+1}}{(l+1)(k-l)!}$$
and then we can reindex so that we start at $k = 1$ like so:
$$e^x\log(1+x) = \sum_{k = 1}^\infty \sum_{l = 0}^{k - 1} \frac{(-1)^{k-1}x^k}{(l+1)(k-l-1)!}$$
and then reindex again so the inner sum to start at $l = 1$ to get the form the other commenters are getting:
$$\boxed{e^x\log(1+x) = \sum_{k = 1}^\infty \sum_{l = 1}^k \frac{(-1)^{k-1}x^k}{l(k-l)!}}$$
So as you can see, the Cauchy product formula can also be used for these series expansions of products, as you originally thought.
Hope this helps!
