How to show that these sequences are summable?

  • #1
zenterix
515
74
Homework Statement
I am reviewing a chapter about uniform convergence in Spivak's Calculus and am reviewing some calculations on infinite sequences and series.

I'd like to know how to show for what values of ##x## the sequences

##\{a_n\}=\left \{(-1)^n\frac{x^{2n+1}}{2n+1}\right \}## with ##n=0,1,2,\ldots##

##\{b_n\}=\left \{(-1)^n\frac{x^n}{n}\right \}## with ##n=1,2,3,\ldots##
Relevant Equations
are summable?
Consider the geometric series ##\sum\limits_{n=1}^\infty r^n##.

This infinite series comes from the sequence ##\{g_n\}=\{1,r,r^2,r^3,\ldots\}##.

If ##|r|\geq 1## then ##\lim\limits_{n\to\infty} g_n=\infty\neq 0##. This limit shows that the geometric series does not converge. That is, ##\{g_n\}## is not summable (when ##|r|>1##).

The geometric series is a power series.

The more general power series centered at ##0## is ##f(x)=\sum\limits_{n=1}^\infty a_nx^n##.

I'd like to be able to prove the following statements that are in Spivak's Calculus, in chapter 24 "Uniform Convergence and Power Series".

We already know that ##\sum\limits_{n=1}^\infty a_nx^n## does not necessarily converge for all ##x##. For example,

$$x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$

converges only for ##|x|\leq 1##, while the power series

$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+\ldots$$

converges only for ##-1<x\leq 1##.

Let ##\{a_n\}=\left \{(-1)^n\frac{x^{2n+1}}{2n+1}\right \}## with ##n=0,1,2,\ldots##.

Let ##\{b_n\}=\left \{(-1)^n\frac{x^n}{n}\right \}## with ##n=1,2,3,\ldots##.

How do we justify when each of these sequences is summable?

If we apply the ratio test to ##\{|a_n|\}## then we have

$$\lim\limits_{n\to\infty} \frac{\frac{x^{2n+3}}{2n+3}}{\frac{x^{2n+1}}{2n+1}}$$

$$=\lim\limits_{n\to\infty} \frac{x^2(2n+1)}{2n+3}$$

$$=x^2$$

which is ##<1## if ##|x|<1##.

By the ratio test, then, if ##|x|<1## then ##\{a_n\}## is absolutely summable, so it is also summable.

The quote above from the book includes ##|x|=1##. Where does this come from? Not the ratio test apparently.

When ##x=1## then we have the sequence ##\{|a_n|\}=\left \{ 1,\frac{1}{3},\frac{1}{5},\frac{1}{7},\ldots \right \}## which is a subsequence of ##\{1/n\}## and the latter is not summable.

The limit comparison test applied to these two sequences shows that

$$\lim\limits_{n\to\infty} \frac{\frac{1}{2n+1}}{\frac{1}{n+1}}=\lim\limits_{n\to\infty}\frac{n+1}{2n+1}=\frac{1}{2}$$

Since ##\frac{1}{2}\neq 0## and ##\{1/n\}## is not summable then neither is ##\{|a_n|\}##.

But this result does not tell us if ##\{a_n\}## is summable or not.

However, we can use Leibniz's theorem: since the sequence is formed by alternating positive and negative terms, the absolute values are decreasing and nonnegative, and the individual terms approach zero, then the sequence is summable.

So yes, we should include ##x=1## as a case in which ##\{a_n\}## is summable.

##x=-1## uses an analogous use of Leibniz's theorem.

Therefore, ##\{a_n\}## is summable for ##|x|\leq 1##.

To show this I used the ratio test and then Leibniz theorem.

Is there some easier way I am missing?

The thing is, the analysis above and any analysis of sequences and series always seems to be a chaotic smorgasboard of these tests without clarity. So I am basically asking if there is an easier method for doing this.

Now let me consider ##\{b_n\}##.
 
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  • #2
Only with experience will you acquire an intuition as to which tests are likely to prove decisive in particular cases.
 

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