0
$\begingroup$

Let c be a fixed number and consider the power series $\displaystyle\sum_{n=1}^ \infty \frac{c^{n-1}}{n} x^{n}$.

a) Determine the convergence radius r for every value of $c \in \mathbb{C}$.

In this task I used the ratio test: $ \mid \frac{a_{n+1}}{a_n} \mid $ this gives me the result $ \mid \frac{n}{n+1}cx \mid$ and this will be $ \mid cx \mid$ as $ n \rightarrow \infty $.

And I know that if $ \mid cx \mid$ $ < 1 $ it will converge. The convergence interval will be $ \frac{-1}{c} < x < \frac{1}{c} $.

To show that this is the convergence interval I would just put the values istead of x and see what happens.

To determine the convergence radius r : $ \frac{\frac{1}{c} -(-\frac{1}{c})}{2} =\frac{1}{c}$.

Is this correct??

b) Let $f$: ]-r,r[ $ \rightarrow \mathbb{C} $ describe the sum function to the power series above.

Show for all real c $\neq $ 0 that $f(x)$ is a strictly increasing function of $x$

Do you have any ideas for this task??

$\endgroup$

1 Answer 1

0
$\begingroup$

Your solution of (a) is correct, although using the formula of Cauchy-Hadamard,

$$ R = \frac{1}{\limsup_{n\rightarrow \infty} \sqrt[n]{|a_n|}}$$

for the power series $\sum_n a_n x^n$ would perhaps have been faster.

For (b) you write your series as $c^{-1} \cdot \sum_n \frac{(cx)^n}{n}$ and then calculate the derivative explicitely (term by term) (why is this legitimate?). Show that the result is non-negative.

$\endgroup$
5
  • $\begingroup$ Okay but do you think the answer for a) is just fine like this. In b) can you refer to a theorem or something. $\endgroup$ May 18, 2014 at 11:20
  • 1
    $\begingroup$ I would shorten your solution as follows: For $x \neq 0$, the ratio test $|a_{n+1}/a_n|$ yields $|\frac{n}{n+1} cx| \rightarrow |cx|$, so that the series converges (absolutely) for $|x| < 1/|c|$ and diverges for $|x| > 1/|c|$ (you forgot the absolute value btw). This shows that the radius of convergence is $r= 1/|c|$. You would have to consider the case $c=0$ explicitely. But then $r = \infty$ is trivial. $\endgroup$
    – PhoemueX
    May 18, 2014 at 11:24
  • $\begingroup$ Why is c absolute valued? $\endgroup$ May 18, 2014 at 16:08
  • 1
    $\begingroup$ Because if $c<0$ then your condition $-1/c < x < 1/c$ is impossible to fulfill. But $|xc| < 1$ can be fulfilled, namely for $|x| < 1/|c|$. $\endgroup$
    – PhoemueX
    May 18, 2014 at 16:13
  • $\begingroup$ Okay, now I got it $\endgroup$ May 18, 2014 at 16:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .