find the sum of each of the following infinite series, if no sum, write no sum.

9-3+1-1/3+1/9...

1/2+1+2+4...

S = a/(1/r) where │r│ < 1

in the first,
a = 9, r = -1/3

so sum = 9/(1 - (-1/3)) = 6.75 or 27/4

For the second one, there is no sum, since the series diverges.
notice r = 2, so it does not satisfy the condition for the formula.

Thank you.

I just notice a typo in my formula, should of course be

a/(1-r) where │r│ < 1

I did use it correctly in the calculations

To find the sum of an infinite series, we need to determine if the series converges or diverges. If it converges, we can then find its sum.

1. For the series 9-3+1-1/3+1/9...
We notice that the common ratio between terms is 1/3. To determine if the series converges, we can apply the formula for the sum of an infinite geometric series:

S = a / (1 - r),

where 'S' represents the sum of the series, 'a' is the first term, and 'r' is the common ratio. In this case, 'a' is 9, and 'r' is 1/3.

Substituting the values into the formula, we get:

S = 9 / (1 - 1/3)
S = 9 / (2/3)
S = 9 * 3/2
S = 27/2

Hence, the sum of the series 9-3+1-1/3+1/9... is 27/2.

2. For the series 1/2+1+2+4...
This series does not have a common ratio between terms, so we cannot use the formula for the sum of a geometric series. We observe that each term is obtained by multiplying the previous term by 2, resulting in an exponential growth.

Therefore, this series diverges, which means it does not have a sum.

In conclusion, the sum of the series 9-3+1-1/3+1/9... is 27/2, while the series 1/2+1+2+4... does not have a sum.