1. Nov 3, 2014

MathewsMD

Hi,

I am likely just missing something fundamental here, but I recently just revisited series and am looking over some notes.

In my notes, I have written that if

$\lim_{x \to +\infty} \frac{a_{n+1}}{a_n} = L$

Then $| x - x_o | = 1/L$

But shouldn't the correct expression be $$| x - x_o | = L$$ ?

Why is the radius of convergence 1/L instead of L? I was under the assumption since

$\frac{(x - x_o)^{n+1}} {(x-x_o)^n} = x - x_o$

that this term (absolute value) would be less than and/or (possibly) equal to L? Are my notes inaccurate or is there a reason for R being 1/L?

Last edited: Nov 3, 2014
2. Nov 3, 2014

Stephen Tashi

What does $x$ have to do with $a_n$. Is $a_n$ some function of $x$ ?

Did you mean $lim_{n\to\infty}$ ?

3. Nov 3, 2014

MathewsMD

Yes. Sorry, that's a typo. It is $lim_{n\to\infty}$ as you noted.

4. Nov 3, 2014

mathman

It would be cleare if you started from the beginning. It is hard to guess the original statement of the situation.

5. Nov 3, 2014

Staff: Mentor

This doesn't make much sense. For starters, the limit should be as n changes, not x. More importantly, if L > 1, the series diverges, so it doesn't make any sense to talk about the radius of convergence.

The limit above appears in the Ratio Test. If L > 1, the series diverges. If L < 1, the series converges. If L = 1, the test isn't conclusive.

6. Nov 4, 2014

Stephen Tashi

Apparently the series in question has the n-th term $T_n = a_n (x - x0)^n$. The ratio relevant to radius of convergence is $\frac{T_{n+1}}{T_n} = \frac{a_{n+1}}{a_n} \frac{(x-x_0)^{n+1}} {(x-x_0)^n}$. If the ratio is to be 1 in the limit and the limit of $\frac{ a_{n+1} } {a_n}$ is $L$ as $n$ approaches infinity then you need the ratio $\frac {(x-x_0)^{n+1}} {(x-x0)^n}$ to be $\frac{1}{L}$.

7. Nov 4, 2014

gopher_p

The power series $\sum\limits_{n=0}^\infty a_n(x-x_0)^n$ converges, by the ratio test, when $\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}\right|<1$ and diverges when $\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}\right|>1$.

After a bit of fiddling, you get that the power series converges when $|x-x_0|<1/L$ and diverges when $|x-x_0|>1/L$, where $L=\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$.