Quick principle questions about two very similar problems in alternating series

[vande060]

Homework Statement

test for convergence by alternating series test

1. (∞, n=1) ∑ ((-1)ⁿ⁺¹ n²)/(n³ + 4)

2. (∞, n=1) ∑ (-1)ⁿ sin(pi/n)

Homework Equations

alternating series test

bn = |an|

alternating series is convergent if satisfies

1. b_n+1 ≤ b_n for all n
2. lim _{n--> ∞} b_n =0

The Attempt at a Solution

1. (∞, n=1) ∑ ((-1)ⁿ⁺¹ n²)/(n³ + 4)

b_n = (n²/(n³ + 4)

f(x) = x²)/(x³ + 4)
f'(x) = x(8-x³)/(x³ + 4)²

f'(x) is decreasing x<0 or x>2, which is not for all n, yet my book still says it converges by this test

2. (∞, n=1) ∑ (-1)ⁿ sin(pi/n)

b_n = sin(pi/n)

f(x) = sin(pi/x), which is clearly decreasing for all n≥2, which again is not all n, but my book says it converges by the alternating series test

what do you think? where am I going wrong?

 

[lanedance]

for 1) you're only interested as n gets large, and the magnitude of the terms clearly go to zero as 1/n

for 2) pi/n tends to zero for large n, so sin(pi/n) tends to zero

 

[vande060]

for 1) you're only interested as n gets large, and the magnitude of the terms clearly go to zero as 1/n

for 2) pi/n tends to zero for large n, so sin(pi/n) tends to zero

okay, I understand the limit of both of these go to zero as n approaches infinity, so the second conidtion is met in the test. Just to clarify what I think you are saying about the first condition for the test, if the series increasing for the first n or two, it is okay, as long an when n is large the series is decreasing, right? That seems to be aside from what the test say though, bn+1 ≤ bn, for all n. I am just not sure I understand :(

 

[lanedance]

thats exactly right

say the terms bob around for the first n terms, then you can add them up, say they sum to M. Now only consider the sum of terms greater than n, if they converge to some limit L, then the original series converges to M+L

 

[vande060]

thats exactly right

say the terms bob around for the first n terms, then you can add them up, say they sum to M. Now only consider the sum of terms greater than n, if they converge to some limit L, then the original series converges to M+L

got it thanks