# Sum of infinite series

1. Jan 31, 2010

### astroboy17

I am trying to understand how to calculate the sum of the following
infinite series, can someone help please:

(5/7)2 - (5/7)3 + (5/7)4 - (5/7)5 + ...

The sum of such a series should be given by:

a / (1-r)

But the value of a = 0 (the first term = 0), hence my confusion.

Thanks

2. Jan 31, 2010

### VeeEight

The value of a is 1

3. Jan 31, 2010

### Staff: Mentor

Rewrite the series with (5/7)2 factored out.

(5/7)2(1 - 5/7 + (5/7)2 -+ ...)
Now can you see that a = 1 as VeeEight suggested?

4. Feb 1, 2010

### Rasalhague

If a = 0, all the terms will be 0.

Another point, the formula you quoted gives the limit of the sum of powers from 0 to n as n goes to infinity:

$$\frac{1}{1-r} = \sum_{k=0}^{\infty} \left( -\frac{5}{7} \right)^k = 1 -\frac{5}{7} + \left(\frac{5}{7}\right)^2 - \left(\frac{5}{7}\right)^3 + ...$$

But your sum is missing the first two terms,

$$\left( -\frac{5}{7} \right)^0 = 1, \left( -\frac{5}{7} \right)^1 = -\frac{5}{7}.$$

So what you have is:

$$\sum_{k=2}^{\infty} \left( -\frac{5}{7} \right)^k = \frac{1}{1-r} - 1 - \left( -\frac{5}{7} \right).$$

The general formula for a geometric series where the summation index begins at some value, m, not necessarily 0, is

$$\frac{ar^m}{1-r}, |r| < 1.$$

http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series

5. Feb 1, 2010

### uart

Astroboy, just use a=25/49 and r=-5/7 in the formula you already posted.

6. Feb 1, 2010

### tiny-tim

Hi astroboy17!

A different way of doing it to to say let the sum be S …

then if you divide S by (5/7)2, you get S + 1 - 5/7 …

ie (7/5)2S = S + 2/7 …

carry on from there.

7. Feb 1, 2010

### Rasalhague

Then we can still use the formula

$$\frac{a}{1-r}$$

if we just redefine a as (-5/7)2, so:

$$\sum_{k=2}^{\infty}\left ( -\frac{5}{7} \right )^k = \sum_{k=0}^{\infty} \left(-\frac{5}{7} \right )^2 \left( -\frac{5}{7} \right )^k$$

$$= \left(-\frac{5}{7} \right )^2 \enspace \frac{\sum_{k=2}^{\infty} \left( -\frac{5}{7} \right )^k}{\left(-\frac{5}{7} \right )^2}$$

$$= \frac{\left( -\frac{5}{7} \right )^2}{1-\left( -\frac{5}{7} \right )}.$$