Problem 2. For each of the following series, determine if it converges and prove your answer.(a)∞∑n=1n44n (b)∞∑n=1n!n4 + 3 (c)∞∑n=12nn!
Question
Problem 2. For each of the following series, determine if it converges and prove your answer.(a)∞∑n=1n44n (b)∞∑n=1n!n4 + 3 (c)∞∑n=12nn!
Solution
(a) ∞∑n=1n/4^n
We can use the ratio test to determine if this series converges. The ratio test states that if the limit as n approaches infinity of the absolute value of the (n+1)th term divided by the nth term is less than 1, then the series converges.
For this series, the (n+1)th term divided by the nth term is ((n+1)/4^(n+1))/(n/4^n) = (n+1)/4n.
As n approaches infinity, this ratio approaches 1/4, which is less than 1. Therefore, the series converges.
(b) ∞∑n=1n!/(n^4 + 3)
For this series, we can use the comparison test. The comparison test states that if 0 ≤ a_n ≤ b_n for all n and the series of b_n converges, then the series of a_n also converges.
We can compare this series to the series of n!/n^4. As n approaches infinity, n^4 + 3 is less than or equal to n^4. Therefore, n!/(n^4 + 3) is greater than or equal to n!/n^4.
The series of n!/n^4 diverges because the factorial function grows faster than any polynomial function. Therefore, the series of n!/(n^4 + 3) also diverges.
(c) ∞∑n=12^n/n!
For this series, we can use the ratio test again. The (n+1)th term divided by the nth term is (2^(n+1)/(n+1)!)/(2^n/n!) = 2/(n+1).
As n approaches infinity, this ratio approaches 0, which is less than 1. Therefore, the series converges.
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