Determine whether the sequence converges or diverges

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This discussion focuses on determining the convergence or divergence of various sequences and series, specifically addressing questions related to limits and monotonicity. The sequences in question include {1 + [(-1)^n]/2}, {1 + [(-1)^n]/3n}, and {sin(n)/n}. The forum participants emphasize the importance of applying the limit test for series convergence, particularly for the series \(\sum \frac{4n+2}{4n-2}\), which requires evaluating the limit of the summand as n approaches infinity.

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Hey all,

University for me started last week, however I was unable to attend until now. I just e-mailed my professor and there is a quiz just next week and I have no notes and he will not give them to me. I'm wondering if anyone knows any good online sites that could help me catch up. Here are some of the questions to give you an idea of what I am looking for:

Question 1: Determine whether the sequence converges or diverges. If it converges, find the
limit.

(a) {1 + [(-1)^n]/2 }

(b) {1 + [(-1)^n]/3n }

(c) {sin(n)/n}

Question 2: Determine whether the sequence is increasing, decreasing or not monotonic. Is the sequences bounded? If the sequence is convergent, find its limit.

(a) { (sqrt(n)) / 1 + sqrt(n) }

(b) { 2 + 1/3^n }

Question 3: Determine whether the series is convergent or divergent. If it is convergent, find
its sum.

(a) \sum 4n+2 / 4n - 2

Any advice on some good reference material would be GREATLY appreciated.. Thanks in advance!
 
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For question three, just see what the summand tends to as n tends to infinity. If it doesn't tend to zero, your sum won't converge.
 


Wouldn't both the numerator and denominator go to infinity?
 


Yes, but that doesn't tell us much. For this one, you can rewrite the function as such:

\frac{4n+2}{4n-2} = \frac{4n-2+4}{4n-2} = 1 + \frac{4}{4n-2}

Try evaluating the limit from here.
 


So it is convergent? And it's sum is 1?
 


No, that's not the series. That's just the summand, the term in the series. You do know the limit test, right?

Limit test:
If an does not tend to 0 as n tends to infinity, then \sum a_n diverges.
 


Like I said, haven't been to class yet. I'll try and get some notes tomorrow, thanks for the help though I appreciate it man.
 

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