# Alternating series test

#### Alexmahone

##### Active member
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
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
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

##### Member
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
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
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

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

##### Member

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

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

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

#### Alexmahone

##### Active member
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
That's the point of the counterexample.
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
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
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
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".

#### Plato

##### Well-known member
MHB Math Helper
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
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