Regarding series convergence or divergence

[eric.chung]

I'm trying to see whether a series is convergence or divergence.

∑_(n=1)^∞ =n^8/12n

Can i consider this a harmonic series and say it diverge?

 

[pka]

I'm trying to see whether a series is convergence or divergence.
∑_(n=1)^∞ =n^8/12n
Can i consider this a harmonic series and say it diverge?

Don't you realize that \(\dfrac{n^8}{n}=n^7~?\)

 

[eric.chung]

Don't you realize that \(\dfrac{n^8}{n}=n^7~?\)

so we just disregard the 12 because it doesn't matter. Is this the direct comparison test?

 

[Steven G]

The 12 should be there but then it factors in from of the summation sign anyways.

Do you know what the harmonic series is? You seem to be saying that if any series has an n in the denominator then it diverges. But then all series will diverge! Why? Multiply every series by n/n and then you'll have an n in the denominator and it converges because it is harmonic. Do you accept this? Remember you CAN multiply any series by 1 in the form of n/n putting that n in the denominator. So does all series diverges?

I have one more point to make. n^8/12n = [math]\dfrac{n^8}{12}n = \dfrac{n^9}{12}[/math]. I suspect that you want 12n in the denominator so put parenthesis around them. n^8/(12n). Also it is NOT [math]\Sigma = n^8/(12n) [/math] but rather [math]\Sigma n^8/(12n) = something[/math]
So does your series converge or diverge? Why?