Determine whether the series is convergent o divergent and say what test you used to solve it.
(d) sum n=1 to infinity
(5n)^(3n) / (5^n + 3)^n
(e) sum k=1 to infinity
5 / sqr(2k - 1)
To determine whether the given series (d) is convergent or divergent, we can use the ratio test.
The ratio test is a test that helps us determine the convergence or divergence of a series by examining the limit of the ratio of consecutive terms.
For series (d), let's apply the ratio test:
1. Calculate the absolute value of the ratio of consecutive terms:
|(5(n+1))^(3(n+1)) / (5^(n+1) + 3)^(n+1)| / |(5n)^(3n) / (5^n + 3)^n|
2. Simplify the ratio:
|5^(3(n+1)) * (5^n + 3)^n / (5^(n+1) + 3)^(n+1) * 5n^(3n) / (5^n + 3)^n|
3. Rearrange terms:
|5^3(n+1) * 5^n * 5^n / (5^(n+1) + 3)(5^n + 3)^n * (5^n + 3)^n / 5n^3n|
4. Cancel out like terms:
|5^3(n+1) * 5^n / (5^(n+1) + 3)(5^n + 3)^n * (5^n + 3)^n / 5n|
5. Simplify further:
|5^3 * (5/n+1) * (5^n)^2 / (5 + 3/5^n+1) * (5^n)^2 / 5n|
6. Take the limit as n approaches infinity:
Limit as n approaches infinity of |5^3 * (5/n+1) * (5^n)^2 / (5 + 3/5^n+1) * (5^n)^2 / 5n| =
|5^3 * 0 * ∞ / 5| = 0
Since the limit of the ratio is less than 1 (0 < 1), the series is convergent by the ratio test.
For series (e), to determine convergence or divergence, we can use the comparison test.
The comparison test allows us to compare a given series with a known series to determine if it has the same convergence or divergence.
Let's compare series (e) with the known series 1 / k^(1/2):
1. Take the limit as k approaches infinity of the ratio between the given series and the known series:
Limit as k approaches infinity of (5 / √(2k-1)) / (1 / √(k)) =
5 * √(k) / (√(2k-1))
2. Simplify the ratio by multiplying the numerator and denominator by √(k) to eliminate the square root terms:
5 * (√(k))^2 / (√(2k-1)) * (√(k))^2 =
5k / (√(2k-1)) * (√(k))^2 =
5k / (√(2k-1)) * k =
5k^2 / (√(2k-1))
3. Take the limit as k approaches infinity:
Limit as k approaches infinity of 5k^2 / (√(2k-1)) =
∞ / ∞
Since we have an indeterminate form (∞ / ∞), we need to simplify the expression further.
To do this, we can multiply the numerator and denominator by the conjugate of the denominator:
4. Multiply the numerator and denominator by the conjugate of (√(2k-1)):
(5k^2 / (√(2k-1))) * (√(2k-1)) / (√(2k-1)) =
5k^2 * (√(2k-1)) / (2k-1)
Now, we can take the limit as k approaches infinity:
5. Limit as k approaches infinity of 5k^2 * (√(2k-1)) / (2k-1) = ∞
Since the limit is not finite (convergence), the given series (e) is divergent according to the comparison test.