Partial sums of alternating Harmonic series satisfy Cauchy Criterion?

**aukie** · December 11th 2009, 10:40 PM

Hello

I am a little stuck in seeing how the partial sums of an alternating Harmonic series satisfy the cauchy condition.

i.e. The partial sums are defined as $s_n=\underset{i=0}{\overset{n}{\sum }}\frac{(-1)^i}{i}$

This yields for $n<m$

$\left|s_m-s_n\right|=\left|\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}\right|<\frac{1}{n}$

Its this last inequality I am having trouble verifying. Any ideas?

**tonio** · December 11th 2009, 11:31 PM

Originally Posted by aukie

Hello

I am a little stuck in seeing how the partial sums of an alternating Harmonic series satisfy the cauchy condition.

i.e. The partial sums are defined as $s_n=\underset{i=0}{\overset{n}{\sum }}\frac{(-1)^i}{i}$

This yields for $n<m$

$\left|s_m-s_n\right|=\left|\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}\right|<\frac{1}{n}$

Its this last inequality I am having trouble verifying. Any ideas?

You can try induction on $m\,,\,m>n$ :

First, it is not hard to see that $\left|\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}\right|=\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}$ .

Now we have two cases: either $m = n+2k\,,\,k\in \mathbb{N}$ , or $m=n+2(k+1)\,,\,k\in\mathbb{N}$ .
In the first case it is trivial, whereas in the second case you can use inductive hypothesis not with m-1 but with m-2, since

if $\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m-1}<\frac{1}{n}$ , then $\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm\frac{1}{m-1}\mp \frac{1}{m}\pm\frac{1}{m+1}<\frac{1}{n}\mp\frac{1} {m}\pm\frac{1}{m+1}$

Tonio

**aukie** · December 12th 2009, 01:59 AM

Hello Tonio

In working out why the absolute value can be ommitted and what you meant by trivial. I seen an argument that doesn't require induction. If you're interested: Assuming m = n + k,

$\frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k}$

Group the terms as follows (for even k)

$\frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - (\frac{1}{n+(k-1)} - \frac{1}{n+k})$

(and for odd k)

$\frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - ( \frac{1}{n+k})$

Then we have a series of positive terms subtracted from $\frac{1}{n+1}$ so

$\frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k} < \frac{1}{n+1} < \frac{1}{n}$

Thanks for your input

**tonio** · December 12th 2009, 07:31 AM

Originally Posted by aukie

Hello Tonio

In working out why the absolute value can be ommitted and what you meant by trivial. I seen an argument that doesn't require induction. If you're interested: Assuming m = n + k,

$\frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k}$

Group the terms as follows (for even k)

$\frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - (\frac{1}{n+(k-1)} - \frac{1}{n+k})$

(and for odd k)

$\frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - ( \frac{1}{n+k})$

Then we have a series of positive terms subtracted from $\frac{1}{n+1}$ so

$\frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k} < \frac{1}{n+1} < \frac{1}{n}$

Thanks for your input

I think it works just fine! Yet even in this case we have what's sometimes called "concealed or hidden" induction: it is a weak form and it just could be stressed that what you did actually works no matter what $n$ in $\frac{1}{n+k}$ you used...

Tonio

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