# Thread: Series of positive and negative terms

1. ## Series of positive and negative terms

Hey guys!
Final question....finally! Can anyone help me with it please??

Investigate the convergence of the series
$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}$,$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\frac{3}{2}}}$

Thanks guys!
Jo

2. Originally Posted by simplysparklers
Hey guys!
Final question....finally! Can anyone help me with it please??

Investigate the convergence of the series
$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}$,$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\frac{3}{2}}}$

Thanks guys!
Jo
The alternating series test is your good friend for testing both series ....

p.s. They both converge (the alternating series test is my good friend too).

3. Ok, note that $\displaystyle a_n=\frac{1}{2n-1}$ is a decreasing sequence, and $\displaystyle \lim a_n=0$, thus by Leibniz Test $\displaystyle \sum{a_n\cdot{(-1)^{n-1}}}$ converges

Alternating series test - Wikipedia, the free encyclopedia

You can do the same for the second one, or see that $\displaystyle \sum{\frac{1}{n^{3/2}}}$ converges, thus $\displaystyle \sum{\frac{(-1)^{n-1}}{n^{3/2}}}$ must be convergent

4. Originally Posted by PaulRS
Ok, note that $\displaystyle a_n=\frac{1}{2n-1}$ is a decreasing sequence, and $\displaystyle \lim a_n=0$, thus by Leibniz Test $\displaystyle \sum{a_n\cdot{(-1)^{n-1}}}$ converges

Alternating series test - Wikipedia, the free encyclopedia

You can do the same for the second one, or see that $\displaystyle \sum{\frac{1}{n^{3/2}}}$ converges, thus $\displaystyle \sum{\frac{(-1)^{n-1}}{n^{3/2}}}$ must be convergent
Just to clarify on PaulRS's completely correct statement

If $\displaystyle \sum{|a_n|}$ converges this means that it converges absolutely

and if a series converges absolutely it converges conditionally(with the $\displaystyle (-1)^n$

But the converse is not necassarily true

I will use these two as examples

For the first one we first test for absolute convergence

So we check if $\displaystyle \sum_{n=1}^{\infty}\bigg|\frac{(-1)^{n-1}}{2n-1}\bigg|=\sum_{n=1}^{\infty}\frac{1}{2n-1}$ converges

To this we notice that all the terms are positive and $\displaystyle a{n+1}<a_n$

So the integral test applies

So we set up the integral test

$\displaystyle \int_1^{\infty}\frac{dn}{2n-1}=\frac{1}{2}\ln|2n-1|\bigg|_1^{\infty}=\infty$

and so since the integral diverges the series diverges. So we know that this series is not absolutely convergent

but now we check for conditional convergence. To do this we use the alternating series test which states that if a series of the form $\displaystyle \sum(-1)^na_n$ and $\displaystyle \exists{N}\backepsilon\forall{n>N}\text{ }a_{n+1}<a_n$ which means that if you can pick any number $\displaystyle \in\mathbb{N}$ and after than number $\displaystyle a_n$ is monotonically decreasing ($\displaystyle a_{n+1}<a_n$)

If this applies then we can use the altnerating series test which states

that if $\displaystyle \lim_{n\to\infty}a_n=0$ the series is convergent

if $\displaystyle \lim_{n\to\infty}a_n\ne{0}$ the series is divergent

So since our series $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1}$

Meets this criteria (e.g. $\displaystyle \forall{n}>0\text{ }a_{n+1}<a_n$

the alternating series test applies

So we check to see if $\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{2n-1}=0$

since direct substitution yields no indeterminate forms and its value is 0 we conclude the limit is 0

Therefore this series converges by the alternating series test

That showed that the converse of "if a series converges absolutely it converges conditionally" is untrue

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Now for your other series

we check for absolute convergence

so first we find $\displaystyle \sum{|a_n|}=\sum_{n=1}^{\infty}\bigg|\frac{(-1)^{n-1}}{n^{\frac{3}{2}}}\bigg|=\sum_{n=1}^{\infty}\fra c{1}{n^{\frac{3}{2}}}$

So either seeing that this is a convergent p-series or seeing that

$\displaystyle \int_1^{\infty}\frac{dn}{n^{\frac{3}{2}}}=\frac{-2}{\sqrt{x}}\bigg|_1^{\infty}=2$

so since the integral converges the series converges.

Now since we have shown that the series absolutely converges it MUST by definition converege conditionally
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Remember to follow instructions though. Generally when they ask you to find the convergence they mean conditional not absolute

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