Determine the convergence or divergence of each of the following series. Find th

Question

Determine the convergence or divergence of each of the following series. Find the sum of the convergent series if possible. If the series is a convergent alternating series then determine if the convergence is absolute or conditional.

Determine the convergence or divergence of each of the following series. Find the sum of the convergent series if possible. If the series is a convergent alternating series then determine if the convergence is absolute or conditional. 2n3 - 3 + 6/n4 + 2n3 + 2 sin 1/2/n2 (n + 3)2n/(4n2 - 5)n (-1)n (2n/1 + 3n)n

Explanation / Answer

1. The sequence tends to 2/n. This is a p-series with p = 1, and if p <= 1, the series diverges

2. As sin 1/n tends to 1/n, this tends to 1/n^3     As p > 1, this converges

3. (n+3)^2n = n^2n*(1+3/n)^2n < n^2n *e^6; more particularly, lim n-> inf (n+3)^2n/n^2n = e^6

(4n^2-5)^n = (4n^2)^n(1 - 5/n^2)^n =4^n*n^2n* (1 - 5/n^2)^n

lim n-> inf (1 - 5/n^2)^n = 1, so lim n-> inf (4n^2-5)^n/n^2n = 4^n

Thus, lim n-> inf (n+3)^2n/(4n^2-5)^n = lim n->inf (n+3)^2n/n^2n/((4n^2-5)^n/n^2n)

= e^6/4^n

This has ratio 1/4, so the series converges.

4. As 2n/(3n+1) < 2/3, the absolute value of all terms is less than (2/3)^n, and this series has ratio 2/3 and converges, so the original series converges.

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