Infinite Series Formula

**Tclack** · August 22nd 2010, 11:30 AM

So Here's a few problems where I have to find the nth partial sum:

Given the equation $S_n = a\frac{1-r^{(n+1)}}{1-r}$

The sum of a Geometric series for a sequence $A_n$ such that it follows the general pattern $A_n = ar^k$ starting at n=1. a is a constant, and
|r| < 1
If we let n go towards infinity, then $S_n = \frac{a}{1-r}$

However, before I can begin dealing with infinity, I can't get the general equation for the sum up to the nth term. What am I doing wrong?

The Sequence is: $2, \frac{2}{5^2}, \frac{2}{5^3}, .... , \frac{2}{5^{(k-1)}}$

$S(\frac{2}{5^{(n-1)}}) = 2(\frac{1}{5})^{(n-1)} = 10(\frac{1}{5})^n$

if we let a = 10 and r = 1/5, then from Sn I got
$10\frac{(1-(\frac{1}{5})^{k+1})}{1-\frac{1}{5}}=10\frac{(1-(\frac{1}{5})^k\frac{1}{5}}{\frac{4}{5}}=$ $\frac{25}{2}(\frac{5}{5}-(\frac{1}{5})^k\frac{1}{5})=\frac{5}{2}(5-(\frac{1}{5})^k)$

That's what I got. However the way it is done is:

$\frac{2}{5^{(k-1)}} = 2(\frac{1}{5})^{(k-1)}$
let a = 2 and let r= 1/5

$S(2(\frac{1}{5})^{(k-1)}=2\frac{\(1-(\frac{1}{5})^{k-1+1}}{1-\frac{1}{5}}= 2\frac{\(1-(\frac{1}{5})^{k})}{1-\frac{1}{5}}= 2\frac{\(1-(\frac{1}{5})^{k}}{\frac{4}{5}}=\frac{5}{2}(1-(\frac{1}{5})^{k}$

Manual calculation by hand shows the second method to be the correct one. While I accept it, I can't see what's wrong with the first.

**Vlasev** · August 22nd 2010, 02:02 PM

I think you are just messing up a lot of the coefficients and that is why you are getting different answers! Here is how to do it properly. The geometric sum is:

$\displaystyle \sum_{i=0}^n ar^i = a\frac{r^{n+1}-1}{r-1}$

In your case you have $n = k-1$ , so your sequence will sum to

$\displaystyle a\frac{r^{k}-1}{r-1}$

Now, your term is $2(1/5)^i = 10(1/5)^{i+1}$ but you have changed the power (you have added 1). Hence you are now summing the following

$\displaystyle \sum_{i=0}^{k-1} ar^{i+1} = \sum_{i=1}^{k} ar^{i} = a\frac{r^{k+1}-1}{r-1}-ar^{0} = a\left(\frac{r^{k+1}-1}{r-1}-1\right)$

$\displaystyle = ar\frac{r^{k}-1}{r-1}$

In the first case you have $a = 2$ , $r = 1/5$ and you get

$\displaystyle \frac{5}{2}(1-5^{-k})$

In the second case you have $a = 10$ , $r = 1/5$ and you get (after a slight simplification)

$\displaystyle \frac{5}{2}(1-5^{-k})$

which lo and behold is the same result. The bottom line is that what you have done is correct in the sense that there is more than one way to do this problem. However you are incorrect in your use of the formulas.

The best formula to use is:
$\displaystyle \sum_{k=0}^n a r^k = a\frac{r^{n+1}-1}{r-1}$

Then if you change the power on the r, you need to change the indices of summation. To help you do that, here is an example. It doesn't have the a, but the a is not important, since it is a constant and if you need it you can multiply the result by it!

$\displaystyle \sum_{k=m}^n r^k = \sum_{k=0}^n r^k - \sum_{k=0}^{m-1} r^k$

$\displaystyle = \frac{r^{n+1}-1}{r-1} - \frac{r^{m}-1}{r-1} = \frac{r^{n+1}-r^{m}}{r-1}$

**Tclack** · August 22nd 2010, 02:38 PM

I think that's where the confusion is, so
k=n+1?

so if the kth term is:

$ar^5$

then the formula for the geometric series is:
$a \frac{r^{5}-1}{r-1}$
since k = 5 and n+1=5. Therefore n=4

The formula is NOT:
$a\frac{r^{5+1}-1}{r-1}= a\frac{r^{6}-1}{r-1}$

Am I correct in this thinking?

**Vlasev** · August 22nd 2010, 02:46 PM

It depends heavily on which term you have started on! If you have started at the power equal to 1 and the kth term is $ar^5$ , then you get $a(r^{5}-1)/(r-1)$ . However, if you start at the power of 0 and the kth term is ar^5, then the formula is $a(r^{5+1}-1)/(r-1)$ . If you reread my post, you should see what I mean.

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