How do you calculate the sum of this infinite series?

[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

 

[VeeEight]

The value of a is 1

 

[Mark44]

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?

 

[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

 

[uart]

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

 

[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.

 

[Rasalhague]

Then we can still use the formula

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

if we just redefine a as (-5/7)², so:

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

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

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