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

series in comments

Question

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

anonymous · Answer

$$\sum_{k=1}^{\infty} \frac{ (-1)^{k+1}1\times4\times7...(3k-2) }{ k!2^k }$$

amistre64 · Answer

1.4.7.10.13.16.19.21.24.27
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    1.2.3.4.5.6.7.8.9.10


13.2.2.19.3.7.2.2
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            5

hmm, well it seems at least that the factorial does not consume the top product ....

anonymous · Answer

I tried the ratio test but it came out to be inconclusive. Im not sure if I did it correct though

amistre64 · Answer

(n+1)/n

amistre64 · Answer

$$\frac{ (-1)^{k+2}1\times4\times7...(3(k+1)-2) }{ (k+1)!2^{(k+1)} }~~
\frac{ k!2^k }{ (-1)^{k+1}1\times4\times7...(3k-2) }$$

$$\frac{ (-1)~(3k+1)}{ 2(k+1) }$$

amistre64 · Answer

looks to me to be -3/2

amistre64 · Answer

as the limit of the sequence, but that might have to be absoluted

anonymous · Answer

which test did you use?

amistre64 · Answer

i used the ratio test

since the limit of the sequence itself is not 0, then it would diverge

anonymous · Answer

Oh okay

anonymous · Answer

Thanks a lot! :)

amistre64 · Answer

youre welcome, i cant get the wolf to verify it .... it hates my syntax :)