Assignment 1: Sequences and Series
1. Determine whether each of the following sequences converges. If it does, find the limit:
(a) $a_n=\frac{n}{n+1}$
(b) $b_n=\frac{(-1)^n}{n}$
2. Test the convergence or divergence of the series:
(a) $\sum_{n=1}^{\infty} \frac{1}{n^2}$
(b) $\sum_{n=1}^{\infty} \frac{n}{n+1}$
3. Use the Ratio Test to determine the convergence of:
$\sum_{n=1}^{\infty} \frac{3^n}{n!}$
4. Determine whether the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges absolutely or conditionally.
5. A geometric series has first term 5 and ratio $r=\frac{2}{3}$. Find the sum of the series.
Answer (1)
Sequence a n = n + 1 n converges to 1.
Sequence b n = n ( − 1 ) n converges to 0.
Series ∑ n = 1 ∞ n 2 1 converges.
Series ∑ n = 1 ∞ n + 1 n diverges, series ∑ n = 1 ∞ n ! 3 n converges, series ∑ n = 1 ∞ n ( − 1 ) n + 1 converges conditionally, and the geometric series sums to 15 .
Explanation
Problem Overview We are given several problems related to sequences and series. We will analyze each one step by step, determining convergence, divergence, and sums where applicable.
Sequence 1(a): Convergence and Limit For sequence a n = n + 1 n , we want to find the limit as n approaches infinity. We can rewrite the expression as a n = 1 + n 1 1 . As n → ∞ , n 1 → 0 , so a n → 1 + 0 1 = 1 . Therefore, the sequence converges to 1.
Sequence 1(b): Convergence and Limit For sequence b n = n ( − 1 ) n , we want to find the limit as n approaches infinity. The numerator oscillates between -1 and 1, while the denominator grows without bound. Thus, the limit is 0. Therefore, the sequence converges to 0.
Series 2(a): Convergence For the series ∑ n = 1 ∞ n 2 1 , we can use the p-test. The p-test states that the series ∑ n = 1 ∞ n p 1 converges if 1"> p > 1 and diverges if p ≤ 1 . In this case, p = 2 , which is greater than 1. Therefore, the series converges.
Series 2(b): Divergence For the series ∑ n = 1 ∞ n + 1 n , we can use the divergence test. We need to find the limit of the terms a n = n + 1 n as n approaches infinity. As we found in problem 1(a), lim n → ∞ n + 1 n = 1 . Since the limit is not 0, the series diverges.
Series 3: Ratio Test For the series ∑ n = 1 ∞ n ! 3 n , we use the Ratio Test. We compute the limit L = lim n → ∞ ∣ a n a n + 1 ∣ = lim n → ∞ ∣ ( n + 1 )! 3 n + 1 ⋅ 3 n n ! ∣ = lim n → ∞ n + 1 3 = 0 . Since L = 0 < 1 , the series converges.
Series 4: Conditional Convergence For the series ∑ n = 1 ∞ n ( − 1 ) n + 1 , we first test for absolute convergence by considering the series ∑ n = 1 ∞ ∣ n ( − 1 ) n + 1 ∣ = ∑ n = 1 ∞ n 1 . This is the harmonic series, which is known to diverge. Therefore, the original series does not converge absolutely. Next, we test for conditional convergence using the Alternating Series Test. The terms a n = n 1 are decreasing and lim n → ∞ n 1 = 0 . Therefore, the series converges conditionally.
Series 5: Geometric Series Sum For the geometric series with first term a = 5 and ratio r = 3 2 , we use the formula S = 1 − r a . Since ∣ r ∣ = 3 2 < 1 , the series converges. The sum is S = 1 − 3 2 5 = 3 1 5 = 15 .
Final Results In summary:
(a) The sequence a n = n + 1 n converges to 1.
(b) The sequence b n = n ( − 1 ) n converges to 0.
(a) The series ∑ n = 1 ∞ n 2 1 converges.
(b) The series ∑ n = 1 ∞ n + 1 n diverges.
The series ∑ n = 1 ∞ n ! 3 n converges.
The series ∑ n = 1 ∞ n ( − 1 ) n + 1 converges conditionally.
The sum of the geometric series is 15.
Examples
Understanding sequences and series is crucial in many fields. For example, in finance, the concept of compound interest relies on the properties of geometric series. If you invest a certain amount of money and it grows at a fixed rate each year, the total amount you have after several years can be calculated using the formula for the sum of a geometric series. This helps in predicting the growth of investments and making informed financial decisions. Similarly, in physics, many phenomena can be modeled using series expansions, such as the Taylor series, which allows us to approximate complex functions with simpler polynomials.
