<s>[INST] $\sum _{n=0}^{\infty } \frac{x^n}{n+3}$, sum, area of convergence & center I want to find the 
$$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
now this is how I thought about doing it but I get stuck.
$$\sum _{n=0}^{\infty } x^n=\frac{1}{1-x}$$ given that absolute value of x is less then zero. 
$$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
I keep integrating the result until I get
$$\frac{x^{n+3}}{(n+3) (n+1) (n+2)}=\frac{3 x^2}{4}-\frac{1}{2} x^2 \log (1-x)-\frac{x}{2}+x \log (1-x)-\frac{1}{2} \log (1-x)$$
I thought about multiplying by $$\frac{(n+1) (n+2)}{x^3}$$
However, the answer should be this: $$\sum _{n=0}^{\infty } \frac{x^n}{n+3}=-\frac{\log (1-x)}{x^3}-\frac{1}{x^2}-\frac{1}{2 x}$$
On the interval [-1,1), and $\frac{1}{3}$ if x=0.
Could someone please show me what I did wrong and how the steps should be to get the correct answer?
 [/INST] I approached it in a slightly different way using Taylor expansions.
The sum for $x=0$ is trivial.
For $x \in (-1,1)$
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\dots$$
$$-\log(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\dots$$
$$-\log(1-x)-x-\frac{x^2}{2}=\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\dots$$
$$-\frac{\log(1-x)}{x^3}-\frac{1}{x^2}-\frac{1}{2x}=\frac{1}{3}+\frac{x}{4}+\frac{x^2}{5}+\dots = \sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
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