Determine whether the following series is divergent, conditionally convergent or absolutely convergent.

Question

anonymous · Answer

$$\sum_{n=1}^{∞}\frac{ n+(-1)^n7 }{ 2n^2+1 }$$

anonymous · Answer

according to my calculations, it is not absolutely convergent.

anonymous · Answer

So I need to show the series are convergent or divergent.

amistre64 · Answer

did you try a comparison?  for large n, this is just 1/2n

amistre64 · Answer

err,  |7/2n^2| maybe

anonymous · Answer

But the comparison are only for positive series, or what?

amistre64 · Answer

|an|  is a positive series .... but im just thinking

amistre64 · Answer

do you know what a limit test is by chance?

anonymous · Answer

Yes but only if I have to show it is absolutely convergent.

anonymous · Answer

Yes I know the limit test..

amistre64 · Answer

"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 · Answer

does this sound reasonable for a comparison?
$$\frac{(-7)^n}{2n^2}$$

amistre64 · Answer

$$\frac{7}{2n^2}\frac{2n^2+1}{7+n}$$
$$\frac{7}{7+n}\frac{2n^2+1}{2n^2}$$
$$\frac{7+n-n}{7+n}\frac{2n^2+1}{2n^2}$$
$$lim~(1-\frac{n}{n+7})(1+\frac{1}{2n^2})$$
(1-1)(1+0) = 0

since the limit is not positive .... it diverges

amistre64 · Answer

or at least thats the way im seeing it ....

anonymous · Answer

Okay. Sp there you show that the series are not absolutely convergent..

amistre64 · Answer

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 · Answer

ugh ... i do loathe these things

anonymous · Answer

:)

amistre64 · Answer

just forget everything ive posted so far ... i was trying to recall a few things that im sure i messed up

anonymous · Answer

Thank you anyway.