MHB Convergence Confusion

[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.

 

[Opalg]

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 $|\,x\,| < R.$

 

[ognik]

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 $$\sum_{n=1}^{\infty}n^3(x-s)^n,$$ 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?

 

[ognik]

... $= \Bigl|\frac{a_{n+1}}{a_n}\Bigr|\,|\,x\,|$ 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).

 

[Prove It]

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...)

 

[ognik]

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

 

[Prove It]

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 $\displaystyle \begin{align*} R^{-1} \, \left| x - c \right| < 1 \end{align*}$...