anonymous
  • anonymous
Determine whether the following series is divergent, conditionally convergent or absolutely convergent.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\sum_{n=1}^{∞}\frac{ n+(-1)^n7 }{ 2n^2+1 }\]
anonymous
  • anonymous
according to my calculations, it is not absolutely convergent.
anonymous
  • anonymous
So I need to show the series are convergent or divergent.

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amistre64
  • amistre64
did you try a comparison? for large n, this is just 1/2n
amistre64
  • amistre64
err, |7/2n^2| maybe
anonymous
  • anonymous
But the comparison are only for positive series, or what?
amistre64
  • amistre64
|an| is a positive series .... but im just thinking
amistre64
  • amistre64
do you know what a limit test is by chance?
anonymous
  • anonymous
Yes but only if I have to show it is absolutely convergent.
anonymous
  • anonymous
Yes I know the limit test..
amistre64
  • amistre64
"Fractions involving only polynomials or polynomials under radicals will behave in the same way as the largest power of n will behave", lamars site ...
amistre64
  • amistre64
does this sound reasonable for a comparison? \[\frac{(-7)^n}{2n^2}\]
amistre64
  • amistre64
\[\frac{7}{2n^2}\frac{2n^2+1}{7+n}\] \[\frac{7}{7+n}\frac{2n^2+1}{2n^2}\] \[\frac{7+n-n}{7+n}\frac{2n^2+1}{2n^2}\] \[lim~(1-\frac{n}{n+7})(1+\frac{1}{2n^2})\] (1-1)(1+0) = 0 since the limit is not positive .... it diverges
amistre64
  • amistre64
or at least thats the way im seeing it ....
anonymous
  • anonymous
Okay. Sp there you show that the series are not absolutely convergent..
amistre64
  • amistre64
i cant really be sure that this is correct, but i think that shows that 7/2n^2 is a viable comparison; and since that diverges .... is it smaller than the original one?
amistre64
  • amistre64
ugh ... i do loathe these things
anonymous
  • anonymous
:)
amistre64
  • amistre64
just forget everything ive posted so far ... i was trying to recall a few things that im sure i messed up
anonymous
  • anonymous
Thank you anyway.

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