<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] Let us consider the partial sums.
\begin{align}
s_N &= \sum_{n=1}^N\left(\frac{x^n}n - \frac{x^{n+1}}{n+1}\right)\\
&= \left(x-\frac{x^2}{2}\right) + \left(\frac{x^2}{2} - \frac{x^3}{3}\right) + \left(\frac{x^3}{3} - \frac{x^4}{4}\right) + \text{ ... } + \left(\frac{x^{N}}{N} - \frac{x^{N+1}}{N+1}\right)\\
&= x - \frac{x^{N+1}}{N+1} 
\end{align}
To evaluate the sum, we take the limit $\lim\limits_{N\to\infty}s_N$ and find that the limit is equal to $x$ if $|x|\leq1$ and does not exist otherwise.
This means that the radius of convergence is $1$.
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