Question

Evaluate the geometric series or state that it diverges. one ninth 1 9plus+startfraction 7 over 81 endfraction 7 81plus+startfraction 49 over 729 endfraction 49 729plus+startfraction 343 over 6561 endfraction 343 6561plus+...

Answer 1

The series converges, and its sum is 1/2.

If r > 1, the series is divergent.  If r < 1, the series is convergent.  In our sequence, r, the common ratio we multiply by to get the next term, is 7/9; therefore it is convergent.

To find the sum of a convergent series, we use the formula
a/(1-r), where a is the first term and r is the common ratio.  We then have
1/9÷(1-7/9) = 1/9÷2/9 = 1/9×9/2 = 1/2

Answer 2

Answer:

In the given geometric series, r, the common ratio is 7/9; therefore the series is convergent.  

Step-by-step explanation:

We are given the geometric series:

$\frac{1}{9} + \frac{7}{81} + \frac{49}{729} + \frac{343}{6561} + ...$

The general geometric series is in the for $a, ar, ar^2, ar^3, ...$, where a is the first term and r is the common ratio.

Comparing with the given geometric series, we have, $a = \frac{1}{9} \text{ and }r = \frac{7}{9}$

If r > 1, the series is divergent.  

If r < 1, the series is convergent.

In the given geometric series, r, the common ratio is 7/9; therefore the series is convergent.