Originally Posted by

**0123** I know these are very very basic series(if they are..

) but I have no idea of what they actually are, I found them in my way and need to understand why the result is that.

If I have the ? [from 1 to 6] of (1/(1.3)^ (s/12)) how do I solve it and why?

Thanks for all the help.

This is a finite geometric series:

$\displaystyle

\sum_{s=1}^6 \frac{1}{1.3^{s/12}} = \sum_{s=1}^6 (1/1.3)^{s/12}=\sum_{s=1}^6 ((1/1.3)^{1/12})^{s}

$

Now you could look up the formula for the sum of a finite geometric series or

derive it form the observation that:

$\displaystyle

(1-b) \left[ \sum_{s=0}^n b^s \right] = 1 - b^{n+1}

$

since if you multiply out the brackets on the left all but the first and last

terms in the result cancel out.

So we have:

$\displaystyle

\left[ \sum_{s=0}^n b^s \right] = \frac{1 - b^{n+1}}{1-b}

$

or:

$\displaystyle

\left[ \sum_{s=1}^n b^s \right] +1 = \frac{1 - b^{n+1}}{1-b}

$

$\displaystyle

\left[ \sum_{s=1}^n b^s \right] = \frac{1 - b^{n+1}}{1-b} -1

$

for any $\displaystyle b$ and $\displaystyle n$, and so in particular for our

series, where $\displaystyle n=6$, and $\displaystyle b=(1/1.3)^{1/12}$.

Hence:

$\displaystyle

\sum_{s=1}^6 \frac{1}{1.3^{s/12}} = \sum_{s=1}^6 (1/1.3)^{s/12}=\sum_{s=1}^6 ((1/1.3)^{1/12})^{s} = \frac{1 - ((1/1.3)^{1/12})^{7}}{1-(1/1.3)^{1/12}}-1$$\displaystyle

= \frac{1 - 1/1.3^{7/12}}{1-1/1.3^{1/12}}-1 \approx 5.56187 $

Reality check:

Code:

>
>.. bute force sum of series:
>
>k=1:6;
>S=(1/1.3)^(k/12);
>SS=sum(S)
5.56187
>
>
>.. CaptainBlack's formula for sum of series:
>
>S6 = (1-1/1.3^(7/12)) / (1-1/1.3^(1/12))-1
5.56187
>

RonL