Convergence of a Power Series: Lim Sup Explained

[kathrynag]

Let sum(a_nx^n) be a power series with a_n not zero, and assume L=lim|a_(n+1)/a_n| exists.
a) Show that if L is not zero, then the series converges for all x in (-1/L,1/L).
b) Show that if L=0, then the series onverges for all x in R
c) Show that a) and b) continue to hold if L is replaced by the limit
L'=lim(s_n) where s_n=sup{|a_(k+1)/a_k|:k>=n}
The value L' is called the limit superior or lim sup of the sequence |a_(n+1)/a_n|. It exists iff the sequence is bounded.
d) Show that if |a_(n+1)/a_n| is unbounded, then the original series converges only when x=0

I'm looking at this and I have no clue how to start.
Like for a), I start by assuming L is not zero.
So we have lim|a_(n+1)/a_n|is nonzero. Then I have trouble getting further.

 

[micromass]

Do you know the following convergence test:

If \sum{u_n} is a series of positive real numbers, then

1. If \lim_{n\rightarrow +\infty}{\frac{u_{n+1}}{u_n}}<1, then \sum{u_n} converges.
2. If \lim_{n\rightarrow +\infty}{\frac{u_{n+1}}{u_n}}>1, then \sum{u_n} diverges.

Look it up in your course, it quite likely that you've shown this somewhere. Otherwise, you'll need to show it yourself, it's not that hard...

 

[kathrynag]

yeah I know that. I just looked at that. I tried looking at a_nx^n=b_n
so I looked at Lim|b_n+1/b_n|.
I just got confused with that. I can see the basic ratio test as you described, but when I try this I get confused.

 

[micromass]

Well, naturally, you would want to apply the ratio test on \sum{a_nx^n}. But the ratio test cannot be applied, since the numbers might not be positive. To fix this, you need to apply the ratio test on \sum{|a_nx^n|}. So, what does the ratio test give you in this case?

 

[kathrynag]

Would it be 1 or -1 based on the rato test of |a_n+1/a_n|, since we look at L<1 or L>1 and to get a value of 1, the absolute value is -1 or 1?

 

[micromass]

Would it be 1 or -1 based on the rato test of |a_n+1/a_n|, since we look at L<1 or L>1 and to get a value of 1, the absolute value is -1 or 1?

I'm sorry, I'm not really following... How did you get -1??

 

[kathrynag]

Well I looked at ratio test of \sum{|a_nx^n|}
I guess I was thinking of it being 1 or -1 based on the absolute value

 

[micromass]

What is 1 or -1?? Can you please write out what you do when applying the ratio test? It's very hard to follow...

 

[kathrynag]

\|a_{n+1}x^{n+1}/a_n|}
Then could I do :
|{\frac{a_{n+1}}{a_n}}||{\frac{x^{n+1}}{x^n}}|

 

[kathrynag]

If <br /> {\frac{a_{n+1}}{a_n}}<br /> is convergent, then <br /> {\frac{a_{n+1}}{a_n}}<br /><1

 

[kathrynag]

so <br /> -1&lt;{\frac{a_{n+1}}{a_n}}<br /><1

 

[micromass]

You should really watch your notation. There's a limit missing there. A lot of mathematicians are quite sensitive to this!

So you obtain \lim_{n\rightarrow +\infty}{\frac{|a_{n+1}x^{n+1}|}{|a_nx^n|}

Now, what does the ratio test say about this limit? (Just repeat the formulation of the ratio test for u_n=|a_nx^n|...)

Furthermore, you can simplify the limit by cancelling a factor xⁿ...

 

[kathrynag]

converges if limit less than 1
-1<{\frac{a_{n+1}x^{n+1}}{a_nx^n}<1
So converges on (-1/L,1/L)

 

[kathrynag]

Ok I think a) and b) make sense now.
Now for c) I look at sup, I see? I guess I would deal with a similar process, but am a little unsure on dealing with the sup.

 

[micromass]

Well, you could show that the ratio test still holds with limsup instead of lim...

 

[kathrynag]

Ok I see!

 

[kathrynag]

Ok I think I'm getting a little stuck on calculating sup. I know sup=smallest upper bound, but how does that work for sup {\frac{|a_{n+1}x^{n+1}|}{|a_nx^n|}?