Series Help...

**larson** · April 20th 2008, 09:19 PM

This question won't take long to answer but I just don't get it...

$S_{60} \approx 14.902$ . Use this result to find upper and lower bounds for S.

And the answer (which is in the back of the book) gives...
$14.902 < S < 14.918$

How did they get this?

**Jhevon** · April 20th 2008, 09:22 PM

Originally Posted by larson

This question won't take long to answer but I just don't get it...

$S_{60} \approx 14.902$ . Use this result to find upper and lower bounds for S.

And the answer (which is in the back of the book) gives...
$14.902 < S < 14.918$

How did they get this?

What?! what is S? you need to give us more information

**larson** · April 20th 2008, 09:24 PM

Originally Posted by Jhevon

What?! what is S? you need to give us more information

Sorry, thats why I was confused also! But then I guess it says refer to example 7.

Example 7 is...
$1 + 2 + 3 + 4 + 5 - \frac {1}{6} + \frac{1}{7} - \frac{1}{8} + \frac{1}{9} - ...$

Which they then group up like this...

$(1 + 2 + 3 + 4 + 5) - (\frac{1}{6} - \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ...)$

**Moo** · April 21st 2008, 12:17 AM

Hello,

I've got doubts on it...

It seems that your series is :

$S_k=1+2+\dots+\frac{k}{2}-(\frac{1}{k/2+1}+\dots+\frac{1}{k})$

Unless I typed it wrong, $S_{60}$ is not 14 etc

**flyingsquirrel** · April 21st 2008, 10:57 AM

Hi

I thinks it is $S_n=1+2+3+4+5-\sum_{k=6}^{n}\frac{(-1)^k}{k}$

As $\sum\frac{(-1)^k}{k}$ is alternating, $\left| \sum_{k=n+1}^{\infty}\frac{(-1)^k}{k} \right| \leq \left|\frac{(-1)^{n+1}}{n+1}\right|=\frac{1}{n+1}$ (1) and $\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k}$ has the sign of $\frac{(-1)^{n+1}}{n+1}$ (2)

Substituting $n=60$ in (1) gives $\left| \sum_{k=61}^{\infty}\frac{(-1)^k}{k} \right| \leq \frac{1}{61}$
Then, $<br /> S = \left| 1+2+3+4+5-\sum_{k=6}^{\infty}\frac{(-1)^k}{k}\right|<br /> \leq \left|\underbrace{ 1+2+3+4+5 -\sum_{k=6}^{60}\frac{(-1)^k}{k}}_{\geq 0} \right|+<br /> \left|\sum_{k=61}^{\infty}\frac{(-1)^k}{k}\right|<br />$
Hence, $S\leq S_{60}+\frac{1}{61}=14.918$
We also have $S=S_{60}-\underbrace{\sum_{k=61}^{\infty}\frac{(-1)^k}{k}}_{\leq 0}$ thanks to (2) whence, $S>S_{60}$ .

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