Answer is: option3
\( -1 \le x < 1 \)Solution:
The radius of convergence for the differentiated series is the same as the radius of convergence for the original series.
Thus, given \[ f(x)=\sum_{n=1}^{\infty}\frac{x^n}{n^2} \] with radius of convergence \(1\), the differentiated series is \[ f'(x)=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n^2} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{n} \]
This series converges for
\[
-1 Checking endpoints:
At \(x=-1\):
\[
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}
\]
which is the convergent alternating harmonic series.
At \(x=1\):
\[
\sum_{n=1}^{\infty}\frac{1}{n}
\]
which is the divergent harmonic series.
Therefore, the interval of convergence is
\[
-1\le x<1
\]
