Asked by Lina
Use the geometric series test to find whether the given series converges or diverges. If it converges, find its sum. 3)
_
Q
k = 1
7
k + 1
9
k - 1
_
Q
k = 1
7
k + 1
9
k - 1
Answers
There are no human answers yet.
Answered by
GPT 3.5
To determine if the given series converges or diverges, we need to check if the common ratio, r, falls within the range of -1 < r < 1.
The given series is:
∑ (7k+1 / 9k-1)
Let's find the value of r:
r = (7(k+1) / 9(k-1)) / (7k / 9k) = (7k + 7) / (9k - 9) * (9k) / (7k)
r = (7k + 7) / (7k - 9)
Now let's check if -1 < r < 1:
-1 < (7k + 7) / (7k - 9) < 1
Next, we can solve for the lower limit:
-1 < (7k + 7) / (7k - 9)
-7k + 9 < 7k + 7
-14 < 14k
-1 < k
So, the lower limit for the common ratio k is k > -1.
Now let's solve for the upper limit:
(7k + 7) / (7k - 9) < 1
7k + 7 < 7k - 9
16 < -9
This inequality is not possible, so there is no upper limit for the common ratio k.
Therefore, the given series diverges.
The given series is:
∑ (7k+1 / 9k-1)
Let's find the value of r:
r = (7(k+1) / 9(k-1)) / (7k / 9k) = (7k + 7) / (9k - 9) * (9k) / (7k)
r = (7k + 7) / (7k - 9)
Now let's check if -1 < r < 1:
-1 < (7k + 7) / (7k - 9) < 1
Next, we can solve for the lower limit:
-1 < (7k + 7) / (7k - 9)
-7k + 9 < 7k + 7
-14 < 14k
-1 < k
So, the lower limit for the common ratio k is k > -1.
Now let's solve for the upper limit:
(7k + 7) / (7k - 9) < 1
7k + 7 < 7k - 9
16 < -9
This inequality is not possible, so there is no upper limit for the common ratio k.
Therefore, the given series diverges.
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