Radius of convergence derivation

[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?

 

[Stephen Tashi]

I have written that if

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

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

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

 

[MathewsMD]

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

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

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

 

[mathman]

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

 

[Mark44]

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 $| x - x_o | = 1/L$

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.

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

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

 

[Stephen Tashi]

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

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 $\frac{1}{L}$.

 

[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 $L=\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$.