Power series, derivatives, integrals, and different intervals of convergence

I am working on the question below. It involves finding three different power series that meet certain conditions.

(a) Find a power series $\sum_{n=0}^{\infty} a_nx^{n}$ that has a different interval of convergence from that of $\sum_{n=0}^{\infty} na_nx^{n-1}$

(b) Find a power series $\sum_{n=0}^{\infty} a_nx^{n}$ such that it has the same radius of convergence with its derivative, but not the same interval of convergence.

(c) Find a power series $\sum_{n=0}^{\infty} a_nx^{n}$ such that its interval of convergence is different from the interval of convergence of its integral.

Here is how I understand how taking the derivative and integral of a convergent power series $\sum_{n=0}^{\infty} a_nx^{n}$ works: So, for (a), I think I would need a power series that: (1) has a radius of convergence of 0, and (2) has a derivative with a radius of convergence $R>0$.

For (b) and (c), I am unsure of what conditions the power series needs to have. How can I control the interval of convergence? Does the initial power series need to be divergent? Then the derivative or integral is convergent and thus has an interval of converence.

I am looking for examples of these power series and help and input in understanding this as well. Thanks very much.

• The issue of convergence at the points $x = a \pm R$ is independent of the convergence within the interval $(a - R, a + R)$. – Sammy Black Mar 28 '13 at 17:50
• A familiar example of this phenomenon: the MacLaurin series for $(1 + x)^{-1}$ diverges at both $x = \pm 1$; whereas, the series for $\log(1 + x)$ converges at $x = 1$. – Sammy Black Mar 28 '13 at 17:55

The radius $R$ of convergence of a power series and its derivative are the same. But the behaviour at the boundary (points with $|x| = R$) might differ. For example, consider $f(x) = \sum_{n \ge 1} \frac{x^n}{n}$. Then $R = 1$, and we can check the series converges for $x = -1$. But $f'(x) = \sum_{n \ge 0} x^n$ diverges for $x = -1$.
As an aside, you may also note that although the series $\sum_{n \ge 0} (-1)^n$ diverges, the limit $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} \frac{1}{1-x}= \frac{1}{2}$ is finite.
I’d say: Take $a_n = \frac{{(-1)}^n}{n}$. I hope I’m not overlooking something.