# Another convergence. An interval of convergence.

• May 19th 2013, 09:22 AM
MathsforNewbs
Another convergence. An interval of convergence.
I have to find, for all value of constant $a>0$, the interval of converence of $\sum_{n=0}^{\infty}\frac{x^{n}}{1+a^{n}}$.
How do I do? By using the Ratio test I get
$\lim_{n \to \infty}\frac{\left (\frac{x^{n+1}}{1+a^{n+1}} \right )}{\left (\frac{x^{n}}{1+a^{n}} \right )}=\lim_{n \to \infty} \frac{x^{n+1}}{x^{n}}\frac{1+a^{n}}{1+a^{n+1}}= \lim_{n \to \infty}x\frac{1+a^{n}}{1+a^{n+1}}$
Now I will use the rule of Hopital to make this fraction 'simpler'.
$\lim_{n \to \infty}x\frac{1+a^{n}}{1+a^{n+1}}=x\lim_{n \to \infty}\frac{\frac{\mathrm{d} \left (1+a^{n} \right )}{\mathrm{d} n}}{\frac{\mathrm{d} \left (1+a^{n+1} \right )}{\mathrm{d} n}}=...=x\lim_{n \to \infty}a^{-1}=\frac{x}{a}$
According to the Ratio test, it states that this series is convergent if $\frac{x}{a}<1$, that is $x. Divergent if $\frac{x}{a}>1$, that is $x>a$. If $x=a$, it has not result. If we test it by inserting the value of $a = 1$ on this series, it would be divergent according to Wolfram Alpha. I don't understand this part because this problem clearly states the value of a should be greater than 0, so $a = 1 > 0$ should also be convergent.

The other hand if we see it converges if $x where $a>0$, so I am not really sure I could say it would be $0. If it's true, the interval of convergent would be like $x \in ]0, a]$ as $a > 0$.
If I did it wrong, please tell me what to do.
• May 19th 2013, 12:45 PM
emakarov
Re: Another convergence. An interval of convergence.
Sorry, I won't analyze your solution and will just write what I think. According to this page, the radius of convergence is the reciprocal of $\lim_{n\to\infty}\left| \frac{1+a^n}{1+a^{n+1}} \right|$. If $a\le1$, the limit is 1, so the radius is 1. If a > 1, the limit is 1 / a, so the radius is a.

Now, what happens when |x| equals the radius? If a < 1 and x = 1, the series diverges by the integral test. If a = x = 1, the series clearly diverges. If a > 1 and x = a, again the series diverges by the integral test. It is left to check when x equals negative radius.
• May 20th 2013, 06:06 AM
MathsforNewbs
Re: Another convergence. An interval of convergence.
Thanks. Didn't think about it before. I have tried to check it and you were right about the values of a and r.
Generally |x| should be lesser than r, so that the series converges. If greater than r, it diverges.
The other hand |x|/a < 1 the series converges which means |x|<a or -a < x < a. So what do I have to answer more presicely?
• May 20th 2013, 06:47 AM
Prove It
Re: Another convergence. An interval of convergence.
You have to realise that there are infinitely many possible series that as written, your expression represents (as r and a are freely chosen). Luckily the ratio test can test all but TWO of these cases, where the ratio's limit is 1. So these two series where the ratio's limit is 1 needs to be tested with some other test as Emakarov has done :)
• May 20th 2013, 10:58 AM
emakarov
Re: Another convergence. An interval of convergence.
Quote:

Originally Posted by MathsforNewbs
The other hand |x|/a < 1 the series converges which means |x|<a or -a < x < a. So what do I have to answer more presicely?

As I said in my previous post, the radius of convergence is not always $a$. This is because $\lim_{n\to\infty} \left| x\frac{1+a^n}{1+a^{n+1}} \right|$ is not always |x| / a. And this is because one of the conditions of the l'Hôpital's rule is that the numerator and the denominator must both tend to 0 or $\pm\infty$. But when a < 1, $1+a^n$ and $1+a^{n+1}$ tend to 1, so the l'Hôpital's rule does not apply. When a < 1, the limit above is |x|, so the radius of convergence is 1. Altogether, the radius is $a$ when a > 1 and 1 when a ≤ 1.

Also, it is easier to see that the interval of convergence is open in all cases not by the integral test, as I said in the previous post, but because when x equals one of the interval's edges, the terms of the series don't tend to zero.