∑∞n=2(−1)n+log(1+np)√n−sinn
A) For which p∈R is the series convergent?
B) For which p∈R is the series divergent, and what is the sum?
C) Is the series absolutely convergent for any p∈R?
I don't even know how to start, so far when I had a problem with parameter I could solve by using some tests. But in this case I can't see any tests I would be able to use directly.
Answer
A-B) For p≤0
∞∑n=2(−1)n+log(1+np)√n−sinn=∞∑n=2(−1)n√n−sinn+∞∑n=2log(1+np)√n−sinn ,
the first term converges by Leibniz criterion and the second term can be written as
∞∑n=2log(1+np)√n−sinn=∞∑n=2np+O(n2p)√n−sinn ,
which conveges for p<−12 and diverges for −12≤p≤0.
For p>0 the numerator is positive so the series has only positive terms:
you obtain divergence by comparison criterion
(−1)n+log(1+np)√n−sinn≥log(1+np)−1√n+1≥C√n+1 .
C) Absolute convergence never occurs since for any p
|(−1)n+log(1+np)√n−sinn|
has a subsequence whose sum is divergent i.e.
1+log(1+(2n)p)√2n−sin2n=1√2n−sin2n+log(1+(2n)p)√2n−sin2n≥1√2n−sin2n≥1√2n+1 .
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