Convergence of Alternating Series with Tricky Terms

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SUMMARY

The forum discussion focuses on testing the convergence of two specific series using the Alternating Series test. The first series, sum of (-1)^n * n/(ln n), diverges because the limit of n/(ln n) approaches infinity, failing the test's requirement that the terms must approach zero. The second series, sum of [sin(n*pi/2)]/n!, can be analyzed by recognizing its relationship to the Taylor series for sine, leading to a convergent form of 1/sqrt(2) * sum of 1/(2n+1)!. The discussion emphasizes the importance of recognizing the behavior of series terms and applying the Alternating Series test correctly.

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Gauss177
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Homework Statement


Test the series for convergence or divergence using the Alternating Series test:

1. sum of (-1)^n * n/(ln n)

2. sum of [sin(n*pi/2)]/n!

Homework Equations




The Attempt at a Solution


1. lim of n/(ln n) goes to infinity (as n->infinity), so it can't satisfy the Alternating Series test. Then if I take limit of the entire sum, I can't figure out what it comes to. I think the limit doesn't exist so the whole thing is divergent, but I'm not sure how to get it.

2. No clue on this one, it's not in the "proper format" as the examples I've seen.

Thanks
 
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1. See if the taylor series for ln x helps.

2. Consider the terms of this series in pairs, n=0,1. What do you notice with the sin part?

EDIT: More clues, because i want to be generous :D

Whats zero divided by anything (other than zero)? You should be able to get that sum to now look like

\frac{1}{\sqrt{2}} \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} Does that converge :)?
 
Last edited:
1. One of the first things you learn about series is that if an does NOT go to 0, then \Sigma a_n does NOT converge.

2. Is this an alternating series? What is the "alternating series test"?
 

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