# Math Help - series..?

1. ## series..?

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.

2. 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.
I'm guessing this is the problem you're trying to solve:

$\sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}$

(I replaced "s" with "k" to make it easier to see.)

If I'm correct, then here's how you would solve it:

First, let $S_6=\sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}$

[1] $S_6=\left(\frac{1}{1.3}\right)^\frac{1}{12}+\left( \frac{1}{1.3}\right)^\frac{2}{12}+\left(\frac{1}{1 .3}\right)^\frac{3}{12}+\left(\frac{1}{1.3}\right) ^\frac{4}{12}+\left(\frac{1}{1.3}\right)^\frac{5}{ 12}+\left(\frac{1}{1.3}\right)^\frac{6}{12}$

Now multiply $S_6$ by a negative $\left(\frac{1}{1.3}\right)^\frac{1}{12}$. This will add 1/12 to the exponent of every term on the right side of the equation, to get:

[2] $-\left(\frac{1}{1.3}\right)^\frac{1}{12}S_6=-\left(\frac{1}{1.3}\right)^\frac{2}{12}-\left(\frac{1}{1.3}\right)^\frac{3}{12}-\left(\frac{1}{1.3}\right)^\frac{4}{12}-\left(\frac{1}{1.3}\right)^\frac{5}{12}-\left(\frac{1}{1.3}\right)^\frac{6}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}$

Now we can add both sides of equations [1] and [2] to get, which elliminates everything but the 1st and 7th terms:

$S_6-\left(\frac{1}{1.3}\right)^\frac{1}{12}S_6=\left(\ frac{1}{1.3}\right)^\frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}$

$\left(1-\left(\frac{1}{1.3}\right)^\frac{1}{12}\right)S_6= \left(\frac{1}{1.3}\right)^\frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}$

$S_6=\frac{1}{1-\left(\frac{1}{1.3}\right)^\frac{1}{12}}\left(\lef t(\frac{1}{1.3}\right)^\frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}\right)\app rox 5.56186547919$

So instead of calculating what all of the terms equalled, you need only to find what these equal to find the total sum. Hope that helps.

----------------------------

Thank you CaptainBlack for checking my work.

3. 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:

$
\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:

$
(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:

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

or:

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

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

for any $b$ and $n$, and so in particular for our
series, where $n=6$, and $b=(1/1.3)^{1/12}$.

Hence:

$
\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$
$
= \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

4. Originally Posted by ecMathGeek
$S_6=\frac{13}{3}\left(\left(\frac{1}{1.3}\right)^\ frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}\right)$
Reality check:

Code:
>S6 = (13/3)*((1/1.3)^(1/12)-(1/1.3)^(7/12))
0.521227
which can't be right as every term in the sum is positive, and the first one
is bigger than this.

So there must be a mistake in your derivation of the result (or in my

Checking through your working (as if done correctly it should give the
correct answer) I can't see where it has gone wrong.

RonL

5. Originally Posted by ecMathGeek
I'm guessing this is the problem you're trying to solve:

$\sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}$

(I replaced "s" with "k" to make it easier to see.)

If I'm correct, then here's how you would solve it:

First, let $S_6=\sum^6_{k=1} \left( \frac{1}{1.3} \right)^{k/12}$

[1] $S_6=\left(\frac{1}{1.3}\right)^\frac{1}{12}+\left( \frac{1}{1.3}\right)^\frac{2}{12}+\left(\frac{1}{1 .3}\right)^\frac{3}{12}+\left(\frac{1}{1.3}\right) ^\frac{4}{12}+\left(\frac{1}{1.3}\right)^\frac{5}{ 12}+\left(\frac{1}{1.3}\right)^\frac{6}{12}$

Now multiply $S_6$ by a negative $\frac{1}{1.3}$. This will add 1 to the exponent of every term on the right side of the equation, to get: (This is where I made my mistake.)

[2] $-\frac{1}{1.3}S_6=-\left(\frac{1}{1.3}\right)^\frac{2}{12}-\left(\frac{1}{1.3}\right)^\frac{3}{12}-\left(\frac{1}{1.3}\right)^\frac{4}{12}-\left(\frac{1}{1.3}\right)^\frac{5}{12}-\left(\frac{1}{1.3}\right)^\frac{6}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}$

Now we can add both sides of equations [1] and [2] to get, which elliminates everything but the 1st and 7th terms:

$S_6-\frac{1}{1.3}S_6=\left(\frac{1}{1.3}\right)^\frac{ 1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}$

$\frac{3}{13}S_6=\left(\frac{1}{1.3}\right)^\frac{1 }{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}$

$S_6=\frac{13}{3}\left(\left(\frac{1}{1.3}\right)^\ frac{1}{12}-\left(\frac{1}{1.3}\right)^\frac{7}{12}\right)$

So instead of calculating what all of the terms equalled, you need only to find what these equal to find the total sum. Hope that helps.
I found my mistake. I'll go back and correct it.