For what values of k does the series ∑n=1 to infinity (11/(2k^2 + 3))^n converge?
I think this is k < -2 and k > 2.
this is just a geometric series, with
r = 11/(2k^2+1)
so, it converges when r < 1 or
2k^2+1 > 11
k^2 > 5
Hey there! You're on the right track, but let me add a little bit of humor to help you remember the correct answer.
Oh, dear series, you want to converge? Well, let me break it down for you. The expression inside that sum, (11/(2k^2 +3)), plays the leading role here. In order for the series to converge, the absolute value of that expression must be less than 1.
Now, to find the values of k that satisfy this condition, we need to solve the inequality:
|11/(2k^2 + 3)| < 1
Clearing the absolute value, we get:
-1 < 11/(2k^2 + 3) < 1
Now, let's consider the numerator and denominator separately to avoid any unnecessary drama.
-1(2k^2 + 3) < 11 < 2k^2 + 3(earning an Oscar for dramatic expression)
If we simplify both inequalities, we get two separate inequality scenarios to consider:
-2k^2 - 4 < 0 AND 2k^2 - 8 < 0
Simplifying further, we get:
k^2 + 2 > 0 AND k^2 - 4 < 0
Let's focus on the second inequality first (because we like to save the best for last). If we solve it, we find:
-2 < k < 2
But, hold on a second, we're told that k > 2 in the beginning. So, that's a contradiction, and we can ignore it.
Now, let's move to the first inequality. As k^2 + 2 is always greater than 0, the value of k doesn't matter in this inequality.
So, drum roll please... the values for k that make the series converge are simply:
k < -2
And that concludes our little series extravaganza, where even the numbers want to be part of the show.
To determine the values of k for which the series ∑n=1 to infinity (11/(2k^2 + 3))^n converges, we can use the ratio test.
The ratio test states that if the absolute value of the ratio of the (n+1)th term to the nth term approaches a value less than 1 as n approaches infinity, then the series converges.
Let's apply the ratio test to the given series:
∑n=1 to infinity (11/(2k^2 + 3))^n
First, let's calculate the ratio of the (n+1)th term to the nth term:
(ratio) = |(11/(2k^2 + 3))^(n+1)| / |(11/(2k^2 + 3))^n|
Simplifying the expression by canceling out the terms within the absolute values:
(ratio) = |(11/(2k^2 + 3))^(n+1) * ((2k^2 + 3)/11)^n|
(ratio) = (11/(2k^2 + 3)) * (2k^2 + 3)/11
(ratio) = (2k^2 + 3)/(2k^2 + 3)
(ratio) = 1
Since the value of the ratio is always 1, the ratio test does not provide any useful information for convergence. Therefore, we need to explore further.
To analyze the convergence of the series, we need to consider the behavior of the expression (11/(2k^2 + 3))^n. Recall that for a series to converge, the terms must approach zero as n approaches infinity.
The expression (11/(2k^2 + 3))^n approaches zero if and only if the absolute value of 11/(2k^2 + 3) is less than 1:
|11/(2k^2 + 3)| < 1
Considering both the positive and negative cases, we can write the following inequalities:
11/(2k^2 + 3) < 1 and 11/(2k^2 + 3) > -1
Simplifying each inequality:
For 11/(2k^2 + 3) < 1:
11 < 2k^2 + 3
8 < 2k^2
4 < k^2
-2 < k < 2
For 11/(2k^2 + 3) > -1:
-11 < 2k^2 + 3
-14 < 2k^2
-7 < k^2
-√7 < k < √7
By taking the intersection of the solutions from both inequalities, we can conclude that the series ∑n=1 to infinity (11/(2k^2 + 3))^n converges for k values between -2 and 2:
-2 < k < 2
Hence, your initial intuition was correct. The series converges for k values less than -2 and k values greater than 2.
To determine the values of k for which the series ∑n=1 to ∞ (11/(2k^2 + 3))^n converges, we can use the ratio test.
According to the ratio test, for a series ∑a_n to converge, the limit of the absolute value of the ratio of consecutive terms, lim(n->∞) |a_(n+1)/a_n|, must be less than 1.
Let's apply the ratio test to our series:
| (11/(2k^2 + 3))^(n+1) / (11/(2k^2 + 3))^n |
Simplifying the expression gives us:
| (11/(2k^2 + 3)) / (1) |
= | 11 / (2k^2 + 3) |
We need to find the values of k for which this limit is less than 1:
lim(n->∞) | 11 / (2k^2 + 3) | < 1
Taking the limit of the absolute value of the expression gives us:
| 11 / (2k^2 + 3) |
< 1
Now, let's analyze the inequality:
| 11 / (2k^2 + 3) | < 1
Since the absolute value of a positive number is always positive, we can remove the absolute value symbol:
11 / (2k^2 + 3) < 1
Simplifying the inequality gives us:
11 < 2k^2 + 3
Subtracting 3 from both sides:
8 < 2k^2
Dividing both sides by 2:
4 < k^2
Taking the square root of both sides:
√4 < |k|
Therefore, the possible values for k are k > 2 and k < -2.
In conclusion, the series ∑n=1 to ∞ (11/(2k^2 + 3))^n converges for k > 2 and k < -2.