Hello,
I don't know what test to use for this series:
Determine the sum of the following series:
inf E n=1 (2^n + 9^n) / 12^n
thank you!
To determine the sum of this series, you can use the concept of geometric series. A geometric series is a sequence of numbers in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.
In your series, the common ratio can be found by dividing the next term by the previous term. Let's calculate it:
(common ratio) = (term n+1) / (term n)
To simplify the expression, let's isolate each term:
term n = (2^n + 9^n) / 12^n
term n+1 = (2^(n+1) + 9^(n+1)) / 12^(n+1)
Now, let's calculate the common ratio using these terms:
(common ratio) = (2^(n+1) + 9^(n+1)) / 12^(n+1) ÷ (2^n + 9^n) / 12^n
Next, we can simplify the expression by multiplying the numerator by (12^n):
(common ratio) = ((2^(n+1) + 9^(n+1)) / 12^(n+1)) * (12^n / (2^n + 9^n))
Simplifying further:
(common ratio) = (2^(n+1) * 12^n + 9^(n+1) * 12^n) / (12^(n+1) * (2^n + 9^n))
Now, let's simplify the numerator by factoring out 12^n:
(common ratio) = (12^n * (2 + 9) * 2^n + 12^n * 9 * 9^n) / (12^(n+1) * (2^n + 9^n))
(common ratio) = (12^n * (11 * 2^n + 9^(n+1))) / (12^(n+1) * (2^n + 9^n))
Now, we can cancel out the common factor of 12^n:
(common ratio) = (11 * 2^n + 9^(n+1)) / (12 * (2^n + 9^n))
After obtaining the common ratio, we need to determine whether it is less than 1 in order for the series to converge. If the common ratio is greater than or equal to 1, the series diverges.
Now, observe that as "n" tends to infinity, the term (11 * 2^n + 9^(n+1)) in the numerator increases much faster than the term (12 * (2^n + 9^n)) in the denominator. This means that the common ratio approaches infinity as n increases, so the series diverges.
Therefore, the sum of this series does not exist (or is considered as infinity).
I hope this explanation helps you understand the process of analyzing the series and determining its sum.