Alternating series test

[Alexmahone]

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.

 

[CaptainBlack]

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

 

[Alexmahone]

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.

 

[Also sprach Zarathustra]

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 $n>1$.

 

[Plato]

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]

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.

 

[Also sprach Zarathustra]

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]

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]

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]

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.

 

[Plato]

$\displaystyle 1-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+\ldots+\frac{1}{2n-1}-\frac{1}{(2n)^2}+\ldots$ 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.

 

[Also sprach Zarathustra]

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 $$ \lim_{n\to\infty}S_{2n}=\infty $$

 

[Plato]

[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]

But $\dfrac{1}{\sqrt{2}+1}<\dfrac{1}{\sqrt{3}-1}$ so it is not decreasing.

That's the point of the counterexample.

 

[Plato]

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]

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]

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]

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]

Agreed. But the question wants us to omit "strictly decreasing", not just "strictly". Please read carefully what I actually wrote.

I give up!

 

[CaptainBlack]

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