Let $\sum_{n=0}^{\infty} \frac{c^nx^n}{n}$ be a power series.
I tried solving using the ratio test
$|\frac{U_n+1}{U_n}|= |\frac{c^{n+1} x^{n+1}}{n+1} \frac{n}{c^nx^n}|= |cx| =\lim_{n \to +\infty} \frac{1}{1+{\frac{1}{n}}} = L = |cx|$
According to ratio test I get :
1.) converges for $|cx| < 1= |x|<\frac{1}{|c|}$
2.) diverges for $|cx| > 1= |x|>\frac{1}{|c|}$
3.) $|cx| = 1 ?$ confirm that in the next step, by checking wheather it's converging at endpoints or not.
so now checking whether series, it's convergent at the endpoints so for $ x = \frac{1}{c}$ and $x=\frac{-1}{c}$.
The result :
At $ x = \frac{1}{c}$ , $\sum_{n=0}^{\infty} \frac{c^n{c}^{-n}}{n}=\sum_{n=0}^{\infty} \frac{1}{n}$ diverges through p test.
At $ x = \frac{-1}{c}$ , $\sum_{n=0}^{\infty} \frac{c^n{-c}^{-n}}{n}=\sum_{n=0}^{\infty} \frac{-1^{n}}{n}$ converges through alternating test
so I can conclude that :
1.) convergent for $x \in [\frac{-1}{c}, \frac{1}{c})$. But statement 1 says " for all c >0 power series converges for atleast one x > 0" is this true?
- rewrite like $x \in (0, \frac{1}{c}) == 0<x< \frac{1}{c}$
2.) and divergent for other x values,otherwise. But statement 2 says " for all c >0 power series diverges for atleast one x > 0" is this true?
-suppose for example I take c = 300.
so, I know x converges for $ x \in (0, \frac{1}{300})$ taking $x = \frac{1}{500}=0.02 > 0 $, which is part of the set.
and diverges for all $ x ≥ \frac{1}{300}>0$
-statement 1 says " for all c >0 power series converges for atleast one x > 0" is this true?
-statement 2 says " for all c >0 power series diverges for atleast one x > 0" is this true?
so is both statements or either statements true?
My answer - I feel both statements should be true
Thanks in advance!
The construction "for all $c > 0$ the ... for at least one $x$ ..." means that we are allowed to specify $x$ dependent on $c$. (This is because $c$ is (universally) quantified before $x$ is (existentially) quantified.) It is sufficient to show, for each possible choice of $c$ that there is a choice of $x$ (which is permitted to depend on $c$) satisfying the part of the sentence that is suppressed in the quote above.
For 1, why not observe that for $c > 0$, one may take $x = \frac{1}{2c}$, which is positive and which satisfies $$ \frac{1}{2c} \in \left[\frac{-1}{|c|}, \frac{1}{|c|} \right) \text{.} $$ You're only required to show (for each choice of $c$) one positive $x$ for which the series converges, so pick any one in the positive half of the interval. Notice that we have exhibited "for each $c > 0$ a specific $x > 0$ where the series converges".
Likewise, for 2, you only have to exhibit one positive $x$ for each choice of $c$. Why not $x = 2/|c|$? Notice that we have exhibited "for each $c > 0$ a specific $x > 0$ where the series diverges".