Question

(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also converges at −R.
(b) Give an example of a power series whose interval of convergence is exactly (−1, 1].

Answer 1

Answer:

a) See the proof below.

b) $\sum \frac{(-x)^n}{n}$

Step-by-step explanation:

Part a

For this case we assume that we have the following series $\sum a)n x^n$ and this series has a finite radius of convergence $R <\infty$ and we assume that $a_n \geq 0$ for all n, this information is given by the problem.

We assume that the series converges at the point $x= R$ since w eknwo that converges, and since converges we can conclude that:

$\sum a)n R^n < \infty$

For this case we need to show that converges also for $x=-R$

So we need to proof that $\sum a_n (-R)^n < \infty$

We can do some algebra and we can rewrite the following expression like this:

$\sum a_n (-R)^n = \sum (-1)^n a)n R^n$ and we see that the last series is alternating.

Since we know that $\sum a_n x^n$ converges then the sequence {$a_n R^n$} must be positive and we need to have $lim_{n\to \infty} a^n R^n = 0$

And then by the alternating series test we can conclude that $\sum a_n (-R)^n$ also converges. And then we conclude that the power series $a_n x^n$ converges for $x=-R$ ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this $\frac{(-x)^n}{n}$

Because the series $\sum \frac{(-x)^n}{n}$ converges to $-ln(1+x)$ when $|x|<!$ using the root test.

But by the properties of the natural log the series diverges at $x=-1$ because $\sum \frac{1}{n} =\infty$ and for $x=1$ we know that converges since $\sum \frac{-1}{n}$ is an alternating series that converges because the expression tends to 0.