MHB SE Class 12 Maths Alternating Series Test

alexmahone
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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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Alexmahone said:
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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CaptainBlack said:
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.
 
Alexmahone said:
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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Alexmahone said:
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?
 
Plato said:
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.
 
Alexmahone said:
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.
 
Also sprach Zarathustra said:
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.
 
Alexmahone said:
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?
 
  • #10
Plato said:
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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  • #11
Alexmahone said:
$\displaystyle 1-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+\ldots+\frac{1}{2n-1}-\frac{1}{(2n)^2}+\ldots$ 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.
 
  • #12
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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  • #13
Also sprach Zarathustra said:
[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.
 
  • #14
Plato said:
But $\dfrac{1}{\sqrt{2}+1}<\dfrac{1}{\sqrt{3}-1}$ so it is not decreasing.

That's the point of the counterexample.
 
  • #15
Alexmahone said:
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.
 
  • #16
Plato said:
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.
 
  • #17
Alexmahone said:
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.
 
  • #18
Plato said:
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.
 
  • #19
Alexmahone said:
Agreed. But the question wants us to omit "strictly decreasing", not just "strictly". Please read carefully what I actually wrote.
I give up!
 
  • #20
Alexmahone said:
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
 
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