Determine convergence or divergence for the following series. State the tests used and justify your answers.
Sum (infinity, n=1) 1/(1+e^-n)
Sum (infinity, n=1) (2*4*6...2n)/n!
Sum (infinity, n=0) (n-6)/n
Sum (infinity, n=0) (n-6)/n!
Sum (infinity, n=0) (100n^14)/4^n
To determine the convergence or divergence of a series, we can use various tests such as the nth term test, comparison test, ratio test, integral test, and the root test. Let's go through each series and apply the appropriate tests to determine their convergence or divergence.
1. Series: ∑ (infinity, n=1) 1/(1+e^-n)
We can use the comparison test to determine the convergence or divergence of this series. Let's compare it with the harmonic series ∑ 1/n.
To do the comparison, we consider that 1/(1+e^-n) > 1/n for all n > 0 because the denominator of the first series is always greater than the denominator of the second series.
Since the harmonic series ∑ 1/n diverges, and the given series is greater than the harmonic series, the given series also diverges.
Conclusion: The series ∑ (infinity, n=1) 1/(1+e^-n) is divergent.
2. Series: ∑ (infinity, n=1) (2*4*6...2n)/n!
This series involves the concept of factorials, and to determine its convergence or divergence, we can use the ratio test.
Let's calculate the ratio:
r = [(2n+2)*(2n+4)/(n+1)!] / [(2n)*(2n+2)/n!]
r = (2n+2)*(2n+4) / (n+1)
As n approaches infinity, the ratio r tends to 4. Therefore, the ratio test tells us that if r < 1, the series converges, and if r > 1, the series diverges.
In this case, r = 4, which is greater than 1. Hence, the series diverges.
Conclusion: The series ∑ (infinity, n=1) (2*4*6...2n)/n! is divergent.
3. Series: ∑ (infinity, n=0) (n-6)/n
To determine the convergence or divergence of this series, we can use the limit comparison test. Let's compare it with the harmonic series ∑ 1/n.
Applying the limit comparison test, let's find the limit:
lim (n→∞) [(n-6)/n] / (1/n)
= lim (n→∞) (n-6) / n * n/1
= lim (n→∞) (n-6)
As n approaches infinity, the limit is infinity.
Since the limit diverges, the series also diverges.
Conclusion: The series ∑ (infinity, n=0) (n-6)/n is divergent.
4. Series: ∑ (infinity, n=0) (n-6)/n!
To determine the convergence or divergence of this series, we can use the ratio test.
Let's calculate the ratio:
r = [(n+1)-6]/(n+1)! / (n-6)/n!
r = (n+1-6)/n+1
As n approaches infinity, the ratio tends to 1. Therefore, the ratio test is inconclusive in this case.
To determine its convergence or divergence, we need to examine the terms of the series. We can see that as n increases, the numerator (n-6) will dominate over the denominator (n!). This suggests that the terms of the series go to infinity as n goes to infinity since the factorial grows much faster than a linear term.
Therefore, the series diverges.
Conclusion: The series ∑ (infinity, n=0) (n-6)/n! is divergent.
5. Series: ∑ (infinity, n=0) (100n^14)/4^n
To determine the convergence or divergence of this series, we can use the ratio test.
Let's calculate the ratio:
r = [(100(n+1)^14)/4^(n+1)] / [(100n^14)/4^n]
r = (100(n+1)^14 * 4^n) / (4^(n+1) * 100n^14)
r = (n+1)^14 / (4n^14)
As n approaches infinity, the ratio tends to 1/4. Since 0 < 1/4 < 1, the ratio test tells us that the series converges.
Conclusion: The series ∑ (infinity, n=0) (100n^14)/4^n converges.