<s>[INST] how to find the radius of convergence ? $\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$ How can i find the radius of convergence ? i dont know where to start i cant use $\frac{a_n}{a_{n+1}}$ test here.
$\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$
the question looks simple but i dont know how to solve it
i got that $R = 0$ but its wrong  because when $x=1$ it is convergent
 [/INST] This is harder, but you can use the ratio test by combining the two fractions under the same denominator.
The $n$th term is equal to: $$\frac{x^n(n+1) - x^{n+1}(n)}{n(n+1)}.$$
Now, for:
$$\lim_{n \to \infty} \left(\frac{x^{n+1}(n+2) - x^{n+2}(n+1)}{(n+1)(n+2)} \cdot \frac{n(n+1)}{x^n(n+1) - x^{n+1}(n)} \right) < 1$$
$$\Rightarrow \lim_{n \to \infty} \left(\frac{x^{n+1}(n+2) - x^{n+2}(n+1)}{x^n(n+1) - x^{n+1}(n)} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \right)<1$$
$$\Rightarrow \frac{x^{n+1} - x^{n+2}}{x^n - x^{n+1}} \cdot 1 \cdot 1 <1$$
$$\Rightarrow |x| <1 \ (*)$$
Therefore, the radius of convergence is $1$.
$(*)$: Going back to the original expression $\frac{x^n}{n} - \frac{x^{n+1}}{n+1}$.
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