Analysis I: Lecture 5

In which we meet three really important tests for convergence.

Our motto for this part of the course is “Tests for convergence are an aid to thinking, not a substitute for thinking”.
We reminded ourselves of the definition of absolute convergence.
Theorem 9 (Absolute convergence implies convergence) If the complex series $\sum\limits_{n=1}^{\infty} a_n$ is absolutely convergent then it is convergent. We proved this first for the case that $a_n$ is real for all $n$ , by splitting into non-negative and negative terms, and then used this to deduce the result even when $a_n$ is complex.
Theorem 10 (Ratio test) Let $\sum\limits_{n=1}^{\infty} a_n$ be a complex series with $a_n \neq 0$ for all $n$ . Suppose that there is some real number $a$ such that $\left| \frac{a_{n+1}}{a_n} \right| \to a$ as $n \to \infty$ . If $a < 1$ , then the series converges absolutely (and so converges). If $a > 1$ , then the series diverges. We saw that this was essentially comparison with a geometric series. (The root test, on Examples Sheet 1, has a similar flavour to the ratio test.)
Theorem 11 (Cauchy condensation test) Let $(a_n)_{n=1}^{\infty}$ be a decreasing sequence of positive numbers. Then $\sum\limits_{n=1}^{\infty} a_n$ converges if and only if $\sum\limits_{n=1}^{\infty} 2^n a_{2^n}$ converges. We proved one direction by collecting terms in blocks corresponding to powers of 2, and we’ll do the other direction similarly next time.

Understanding today’s lecture

Keep adding these new tests and new examples to your series grids. Are you starting to get a feel for which tests are useful for which types of series?
We said that the ratio test is inconclusive if the ratio tends to 1 or if the ratio does not have a limit. Find examples to explore this.
The Cauchy condensation test was for series $\sum\limits_n a_n$ where $(a_n)$ is a decreasing sequence of positive numbers. What happens if the sequence isn’t decreasing, or the terms aren’t all positive? You should get into the habit of investigating whether conditions like this are necessary.
Can you prove the other direction of the Cauchy condensation test?

Preparation for Lecture 6

For which $\alpha$ is the series $\sum\limits_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ convergent?

We have seen that every absolutely convergent series is convergent. Are there any series that are convergent but not absolutely convergent?

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11 Responses to “Analysis I: Lecture 5”

Tom Says:
January 28, 2013 at 4:40 pm
Hi, Dr Neale, I am right in thinking that the conditions we put on the ratio test could be weakened a little? Specifically, I’m thinking that for the argument for the series converging to work, we do not necessarily need (the modulus of the ratio between consecutive terms) to converge, as long as there exists a point (in your sequence) after which it is bounded above by some x which is strictly less than 1.

For example, a series in which terms are alternately 1/2 of the previous term, or 1/4 of the previous term has a non-convergent sequence of ratios, but a ratio test type argument will show it converges.

There might also be a similar thing for the divergent case.
theoremoftheweek Says:
January 28, 2013 at 8:35 pm
Indeed. The idea of the ratio test is “compare with a geometric series”, and there are lots of ways of making that work that aren’t exactly the same as the ratio test. So, like the comparison test, it’s important to be comfortable with the idea of the proof as well as the statement of the test, and as you say comparison with a geometric series is a very useful idea. Thanks for drawing attention to it.
Ville Says:
January 28, 2013 at 9:15 pm
Hi Dr. Neale, I seem to have found some alternative proof to Theorem 9. The idea is to make a cauchy sequence from the partial sums of the “absolute” sequence, and use the triangle inequality to show that the partial sums of the original sequence form a cauchy sequence. In full:

$\sum |a_n|$ converges

therefore partial sums converge
therefore partial sums form cauchy sequence

therefore, for all $\epsilon > 0$ there exists $N$ s.t. for all $m$ , $k > N$

$| \sum\limits_{n=1}^{n=m} |a_n| - \sum\limits_{n=1}^{n=k}|a_n| | < \epsilon$

thus $|\sum\limits_{n=k+1}^{n= m}|a_n| | < \epsilon$

and as for all $n$ , $|a_n| \geq 0$ ,

$\sum\limits_{n=k+1}^{n= m}|a_n| < \epsilon$

thus $|a_{k+1}| + |a_{k+2}| + ... + |a_{m}| < \epsilon$

but by repeated use of triangel inequality,

$|a_{k+1} +a_{k+2} + ... +a_{m} | \leq |a_{k+1}| + |a_{k+2}| + ... + |a_{m}| < \epsilon$

So $| \sum\limits_{n=k+1}^{n=m} a_n | < \epsilon$

and $|\sum\limits_{n=1}^{n=m }(a_n) - \sum\limits_{n=1}^{n=k}( a_n) | < \epsilon$

So partial sums form a cauchy sequence, and therefore converges. So $\sum( a_n)$ converges.

Anyone see any mistakes?

[Edited to put the symbols into LaTeX. --VRN.]
theoremoftheweek Says:
January 28, 2013 at 9:50 pm
I’ve used my magic powers to make the symbols in your comment show up nicely. You should be able to use LaTeX by typing a dollar symbol, then the word ‘latex’, then your LaTeX code, then another dollar symbol.

Hopefully I haven’t introduced any typos — let me know if I have.

I’ll let others think about the argument before I say anything. Any comments, anyone?
Ville Says:
January 29, 2013 at 5:55 pm
Thanks, there’s only one small typo. The second line of sums should read

$| \sum\limits_{n=1}^{n=m} |a_n| - \sum\limits_{n=1}^{n=k}|a_n| | < \epsilon$

[Now fixed. --VRN]
theoremoftheweek Says:
January 29, 2013 at 5:58 pm
It turns out that it’s important that you type the first dollar symbol, then the word `latex’ without a space between it and the dollar symbol, then a space, then the LaTeX!
theoremoftheweek Says:
January 30, 2013 at 10:11 am
This is a nice argument, and is a good illustration of how we can use the General Principle of Convergence in both directions: start with a convergent sequence, deduce that it’s Cauchy, obtain another Cauchy sequence, and deduce that it also converges.

Thanks for sharing it!
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