1. ## Series Help...

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

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

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

How did they get this?

2. Originally Posted by larson
This question won't take long to answer but I just don't get it...

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

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

How did they get this?

3. Originally Posted by Jhevon
Sorry, thats why I was confused also! But then I guess it says refer to example 7.

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

Which they then group up like this...

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

4. Hello,

I've got doubts on it...

It seems that your series is :

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

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

5. Hi

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

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

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