Best guess for the function is
[tex]\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}[/tex]
By the ratio test, the series converges for
[tex]\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1[/tex]
When [tex]x=1[/tex], [tex]f(x)[/tex] is a convergent [tex]p[/tex]-series.
When [tex]x=-1[/tex], [tex]f(x)[/tex] is a convergent alternating series.
So, the interval of convergence for [tex]f(x)[/tex] is the closed interval [tex]\boxed{-1 \le x \le 1}[/tex].
The derivative of [tex]f[/tex] is the series
[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n[/tex]
which also converges for [tex]|x|<1[/tex] by the ratio test:
[tex]\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1[/tex]
When [tex]x=1[/tex], [tex]f'(x)[/tex] becomes the divergent harmonic series.
When [tex]x=-1[/tex], [tex]f'(x)[/tex] is a convergent alternating series.
The interval of convergence for [tex]f'(x)[/tex] is then the closed-open interval [tex]\boxed{-1 \le x < 1}[/tex].
Differentiating [tex]f[/tex] once more gives the series
[tex]\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)[/tex]
The first series is geometric and converges for [tex]|x|<1[/tex], endpoints not included.
The second series is [tex]f'(x)[/tex], which we know converges for [tex]-1\le x<1[/tex].
Putting these intervals together, we see that [tex]f''(x)[/tex] converges only on the open interval [tex]\boxed{-1 < x < 1}[/tex].
