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Alternating series test

Alexmahone

Active member
Jan 26, 2012
268
Alternating series test: If $\{a_n\}$ is positive and strictly decreasing, and $\lim a_n=0$, then $\sum(-1)^n a_n$ converges.

Is the alternating series test still valid if "strictly decreasing" is omitted? Give a proof or counterexample.
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
Alternating series test: If $\{a_n\}$ is positive and strictly decreasing, and $\lim a_n=0$, then $\sum(-1)^n a_n$ converges.

Is the alternating series test still valid if "strictly decreasing" is omitted? Give a proof or counterexample.
Consider the series with terms \(a_n=1/n\) for \(n\) odd and \(2/n\) for \(n\) even.

Now \(\{a_n\}\) is a sequence of positive terms and \( \lim_{n \to \infty}a_n=0 \) but \( \sum (-1)^n a_n\) diverges since the \(k\)-th partial sum is the \(k\)-th partial sum of the alternating harmonic series and the half the \(\lfloor k/2 \rfloor \)-th partial sum of the harmonic series. Hence the sequence of partial sums diverges since the sequence of partial sum of the alternating harmonic series converges and that of the harmonic series diverges.

CB
 
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Alexmahone

Active member
Jan 26, 2012
268
Consider the series with terms \(a_n=-1/n\) for \(n\) odd and \(2/n\) for \(n\) even.
That's not a valid counterexample since $\{a_n\}$ should be positive.
 
Jan 31, 2012
54
Alternating series test: If $\{a_n\}$ is positive and strictly decreasing, and $\lim a_n=0$, then $\sum(-1)^n a_n$ converges.

Is the alternating series test still valid if "strictly decreasing" is omitted? Give a proof or counterexample.
Hmm.

$$\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+...$$


Edit:


Another one:

Take, $ a_n=\frac{1}{\sqrt[3]{n}-(-1)^n}$, with $n>1$.
 
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Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Alternating series test: If $\{a_n\}$ is positive and strictly decreasing, and $\lim a_n=0$, then $\sum(-1)^n a_n$ converges.
Is the alternating series test still valid if "strictly decreasing" is omitted? Give a proof or counterexample.
Let $b_n = \left\lfloor {\dfrac{{n + 1}}{2}} \right\rfloor $ or $1,1,2,2,3,3,\cdots$

Now let $a_n=\dfrac{1}{b_n}$. Does $\sum\limits_n {( - 1)^n a_n } $ converge?
 

Alexmahone

Active member
Jan 26, 2012
268
Let $b_n = \left\lfloor {\dfrac{{n + 1}}{2}} \right\rfloor $ or $1,1,2,2,3,3,\cdots$

Now let $a_n=\dfrac{1}{b_n}$. Does $\sum\limits_n {( - 1)^n a_n } $ converge?
Surely $\displaystyle\frac{1}{1}-\frac{1}{1}+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\ldots$ converges to 0.
 
Jan 31, 2012
54
Surely $\displaystyle\frac{1}{1}-\frac{1}{1}+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\ldots$ converges to 0.
Think again.
 

Alexmahone

Active member
Jan 26, 2012
268
Think again.
Let $\displaystyle\{s_n\}$ be the sequence of partial sums.

$\displaystyle s_{2k-1}=\frac{1}{k}$ and $\displaystyle s_{2k}=0$

Since both the subsequences $\displaystyle\{s_{2k-1}\}$ and $\displaystyle\{s_{2k}\}$ converge to 0, $\displaystyle\{s_n\}$ converges to 0.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Let $\displaystyle\{s_n\}$ be the sequence of partial sums.
$\displaystyle s_{2k-1}=\frac{1}{k}$ and $\displaystyle s_{2k}=0$
Since both the subsequences $\displaystyle\{s_{2k-1}\}$ and $\displaystyle\{s_{2k}\}$ converge to 0, $\displaystyle\{s_n\}$ converges to 0.
That is correct. But can you play with a similar sequence?
 

Alexmahone

Active member
Jan 26, 2012
268
That is correct. But can you play with a similar sequence?
$\displaystyle 1-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+\ldots+\frac{1}{2n-1}-\frac{1}{(2n)^2}+\ldots$ diverges as the sum of the positive terms diverges.
 
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Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
$\displaystyle 1-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+\ldots+\frac{1}{2n-1}-\frac{1}{(2n)^2}+\ldots$ diverges as the sum of the positive terms diverges.
Actually the proof of the alternating test only requires that $a_n\ge a_{n+1}$. So yes you can make that change.
I should have read the OP more closely.
 
Jan 31, 2012
54
Flogging a dead horse...


$$\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+...$$

$$S_{2n}=\sum_{k=2}^{n+1}(\frac{1}{\sqrt{k}-1}-\frac{1}{\sqrt{k}+1})=\sum_{k=2}^{n+1}\frac{2}{k-1}$$

In other words, $S_{2n}$ equals to the first n terms of diverges series $\sum_{k=2}^{\infty}\frac{2}{k-1}$, hence $$ \lim_{n\to\infty}S_{2n}=\infty $$
 
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Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
[h=1]Flogging a dead horse...[/h]
$$\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}+...$$
But $\dfrac{1}{\sqrt{2}+1}<\dfrac{1}{\sqrt{3}-1}$ so it is not decreasing.
 

Alexmahone

Active member
Jan 26, 2012
268
But $\dfrac{1}{\sqrt{2}+1}<\dfrac{1}{\sqrt{3}-1}$ so it is not decreasing.
That's the point of the counterexample.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
That's the point of the counterexample.
Your OP asks if decreasing could be replaced with non-increasings.
That is, can $a_n>a_{n+1}$ can be replaced by $a_n\ge a_{n+1}$?
The answer is yes it can.
The proof of the alternating series test allows that.
I said that I failed to read the OP closely the first time.
Sorry for the confusion.
 

Alexmahone

Active member
Jan 26, 2012
268
Your OP asks if decreasing could be replaced with non-increasings.
That is, can $a_n>a_{n+1}$ can be replaced by $a_n\ge a_{n+1}$?
The answer is yes it can.
The proof of the alternating series test allows that.
I said that I failed to read the OP closely the first time.
Sorry for the confusion.
My OP asks if "strictly decreasing" can be omitted, not if "strictly" can be omitted.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
My OP asks if "strictly decreasing" can be omitted, not if "strictly" can be omitted.
That is exactly what I just said.
If you omit "strictly" from "strictly decreasing" the correct term is "non-increasing".
And the answer is YES it can be omitted with changing the proof.
 

Alexmahone

Active member
Jan 26, 2012
268
If you omit "strictly" from "strictly decreasing" the correct term is "non-increasing".
Agreed. But the question wants us to omit "strictly decreasing", not just "strictly".

Please read carefully what I actually wrote.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Agreed. But the question wants us to omit "strictly decreasing", not just "strictly". Please read carefully what I actually wrote.
I give up!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
That's not a valid counterexample since $\{a_n\}$ should be positive.
Sorry take the signs off, I was writing the terms of the series with the sign alternation.

CB