I am looking for help with a problem. Here is the question I am working on:

Consider the power series

$$\sum_{n=1}^{\infty}(-1)^n \frac{n+1}{n}x^{n}$$ (a) Determine the radius of convergence

(b) If $f(x)$ is the sum of the series, write down the power series expansion at $c=1$ of $f'(x)$ and $\int_{1}^{x}f(t)dt$.

So, starting with (a). I apply the Root Test to derive the following:

$$\limsup_{n \to \infty} |a_nx^n|^{1/n} = |x| \limsup_{n \to \infty} \left | (-1)^n \frac{n+1}{n} \right |^{1/n} = |x| \limsup_{n \to \infty} \left | \frac{n+1}{n} \right |^{1/n} = |x| \cdot 1.$$

So, the power series converges for $|x| < 1$. That is, it has a radius of convergence of $R=1.$ Now for part (b).

What I have is that

$$f(x) = \sum_{n=1}^{\infty}(-1)^n \frac{n+1}{n}x^{n}.$$

Taking the first derivative yields

$$f'(x) = \sum_{n=1}^{\infty} (-1)^n (n+1)x^{n-1} = \sum_{n=1}^{\infty} (-1)^n n x^{n-1} + \sum_{n=1}^{\infty} (-1)^n x^{n-1}$$

Now, I believe that I need to find an expression from this infinite sum and then write out the power series expansion of that expression. My first hurdle was thinking about how to deal with the periodic nature of the power series.

My thought was to write each $\sum_{n=1}^{\infty} (-1)^n n x^{n-1}$ and $\sum_{n=1}^{\infty} (-1)^n x^{n-1}$ as two separate power series, respectively. One accounting for the positive (even) terms of the sum, and one as the negative (odd) terms of the sum. I've stumped myself on that and I was hoping for some input.

I have not completed the second part of (b), that is, taking the integral and then finding the power series expansion, on account of being stuck on the first bit. Thank you for any input and help.

  • $\begingroup$ What did you need help with? The integral can also be performed term by term. $\endgroup$ – copper.hat Apr 5 '13 at 5:39
  • $\begingroup$ The shaded question doesn't ask you to compute a closed form expression for $f$. $\endgroup$ – copper.hat Apr 5 '13 at 5:56

Note that $$f(x) = \sum_{n=1}^{\infty} (-1)^n \dfrac{n+1}n x^n = \sum_{n=1}^{\infty}(-x)^n + \sum_{n=1}^{\infty} \dfrac{(-x)^n}n = -\dfrac{x}{1+x} - \log(1+x)$$


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