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
On
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 $\frac{x^n}{n} - \frac{x^{n+1}}{n+1}$.
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 $1$.