Absolute & conditional convergence

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The discussion focuses on determining the convergence of two series. For the first series, \(\Sigma (-1)^n \frac{n^4}{x^3 + 1}\), the positive series diverges as \(\frac{n^4}{n^3 + 1}\) approaches infinity, leading to uncertainty about the original series' convergence. The second series, \(\Sigma \frac{\sin(x)}{x^2}\), is confirmed to be absolutely convergent by comparison with \(\frac{1}{x^2}\), which converges. The importance of the terms approaching zero for convergence is emphasized, as a series with terms that do not go to zero will diverge. Overall, the key takeaway is the necessity of analyzing term behavior and applying appropriate convergence tests.
magnifik
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Homework Statement


Determine whether the series converges absolutely, conditionally, or not at all.

a) \Sigma (-1)nn4/(x3 + 1)

b) \Sigma sin(x)/x2


Homework Equations





The Attempt at a Solution


a) positive series is n4/n3+1 .. do i do comparison test ??

b) |sin(x)|/x2
compare it with 1/x2 which converges.. so it's absolutely convergent??
 
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You are kind of freely mixing n's and x's here. Are they supposed to be the same? If so, for the first one ask whether the nth term goes to zero. For the second one, yes, it's absolutely convergent.
 
woops, mixing up the n's & x's was a careless mistake
 
for a) the positive series diverges because n^4/n^3 + 1 goes to infinity, but I'm not sure if the original series converges or diverges
 
magnifik said:
for a) the positive series diverges because n^4/n^3 + 1 goes to infinity, but I'm not sure if the original series converges or diverges

A series whose terms don't go to zero diverges no matter what the signs on the terms. Look at the definition of convergence in terms of partial sums.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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