# Thread: Convergence Confusion

1. The Cauchy Ratio test says: If $\lim_{{n}\to{\infty}}\frac{a_{n+1}}{a_n} < 1$ then the series converges. OK.

Now I read that for a power series (of functions of x), the same test also provides the interval of convergence, i.e. If the series converges, then $\lim_{{n}\to{\infty}}\frac{a_{n+1}}{a_n} = {R}^{-1}$ and the interval is -R < x < R

Could someone please explain why this works? Thanks.

2. Originally Posted by ognik
The Cauchy Ratio test says: If $\lim_{{n}\to{\infty}}\frac{a_{n+1}}{a_n} < 1$ then the series converges. OK.

Now I read that for a power series (of functions of x), the same test also provides the interval of convergence, i.e. If the series converges, then $\lim_{{n}\to{\infty}}\frac{a_{n+1}}{a_n} = {R}^{-1}$ and the interval is -R < x < R

Could someone please explain why this works? Thanks.
Both those results need to be stated a bit more carefully.

The Cauchy Ratio test says: If $\color{red}{\sum a_n}$ is a series of positive terms and $\lim_{{n}\to{\infty}}\frac{a_{n+1}}{a_n} < 1$ then the series converges.

If you want to apply the test in a situation (like a power series with $x<0$) where the terms may not all be positive, then you have to take absolute values. The test then says that if $\lim_{{n}\to{\infty}}\Bigl|\frac{a_{n+1}}{a_n}\Bigr| < 1$ then the series converges (absolutely).

For a power series $\sum a_nx^n$, the limit $\lim_{{n}\to{\infty}}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|$ does not necessarily exist. The correct statement of the result is that if that limit exists and is equal to $R^{-1}$, then $R$ is the radius of convergence of the series.

The reason why this works is that you can apply the Cauchy Ratio test to see that the series converges if $\lim_{{n}\to{\infty}}\Bigl|\frac{a_{n+1}x^{n+1}}{a_nx^n}\Bigr| < 1$. But $\Bigl|\frac{a_{n+1}x^{n+1}}{a_nx^n}\Bigr| = \Bigl|\frac{a_{n+1}}{a_n}\Bigr|\,|\,x\,|$ and so $\lim_{{n}\to{\infty}}\Bigl|\frac{a_{n+1}x^{n+1}}{a_nx^n}\Bigr| = R^{-1}|\,x\,|.$ Therefore the series converges if $R^{-1}|\,x\,| < 1$, in other words if $|\,x\,| < R.$

3. Thread Author
Glad to see I was correct about the approach needing the modulus of the ratio, but the point behind R still eludes me.

For example consider a series $\displaystyle \sum_{n=1}^{\infty}n^3(x-s)^n,$ s > 0.

This is convergent and works out to this point: $\lim_{{n}\to{\infty}} |\frac{a_{n+1} }{a_n}| < 1$, from which I get $s-1 < x < s+1$, ie an interval of (s-1, s+1). This I'm happy with - and I didn't need an R.

Using $R^{−1}|x-s|<1 \implies |x-s|<R, \therefore s-R < x < s+R$. To me the R seems add unnecessary complexity, I have to assume that R = 1 to get the actual values, so what am I missing?

4. Thread Author
Originally Posted by Opalg
... $= \Bigl|\frac{a_{n+1}}{a_n}\Bigr|\,|\,x\,|$ and so $\lim_{{n}\to{\infty}}\Bigl|\frac{a_{n+1}x^{n+1}}{a_nx^n}\Bigr| = R^{-1}|\,x\,|.$
Hi Opalg, I just don't follow this step?

Also is what I did wrong? - $\lim_{{n}\to{\infty}} |\frac{a_{n+1} }{a_n}| < 1$, from which I get $s-1 < x < s+1$, ie an interval of (s-1, s+1).

5. Originally Posted by ognik
Hi Opalg, I just don't follow this step?

Also is what I did wrong? - $\lim_{{n}\to{\infty}} |\frac{a_{n+1} }{a_n}| < 1$, from which I get $s-1 < x < s+1$, ie an interval of (s-1, s+1).
You have gotten the INTERVAL of convergence (although you still need to check the endpoints) but you were asked for the RADIUS of convergence.

Since it seems you have gotten \displaystyle \begin{align*} \left| x - s \right| < 1 \end{align*} (I haven't checked myself) then that would mean that the series converges in the circle centred at \displaystyle \begin{align*} (x, y) = (s, 0) \end{align*} and thus the radius of convergence is 1. (I say circle because it could be a complex value...)

6. Thread Author
OK, thanks I was mixing those up, but I still don't follow that step in Opalg's post - where R appears....?

7. Originally Posted by ognik
OK, thanks I was mixing those up, but I still don't follow that step in Opalg's post - where R appears....?
Sigh... You need to get it to a point where \displaystyle \begin{align*} \left| x - c \right| < R \end{align*}, or equivalently \displaystyle \begin{align*} R^{-1} \, \left| x - c \right| < 1 \end{align*}...

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