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P-series. Since the exponent of the denominator is >1, the Series must converge.
And I don't have to do a comparison test because the numerator isn't 1?
Unless you need to prove conditional convergence or absolute convergence
the bottom become larger than the top: 1 --------------- = .00000...000001 1000000...00000 is a very small number.
Nah, I don't have to prove conditional vs. absolute convergence for this, just convergence vs. divergence.
The numerator either being 1 or not being one doesn't really matter to the comparison test. The idea of the comparison test is to prove that a sequence that is smaller diverges or that a sequence that is larger converges. If you find either one of those conditions, you have proven the convergence or divergence of the series you are discussing.
I thought it was the other way around? Proving that the series that's smaller than the known convergent one also converges or the series that's larger than the known divergent one diverges. I might be splitting hairs at this point, though. Thanks.
I am talking about the sequence, not the series. The relationship between the sequence and the series is the fact that if your sequence is anything but zero, you are adding to your series (since the summation of your sequence is your series after all)
When a series is convergent, it means that it's sequence either becomes zero at a point or it increases too slowly to really add to the series. (.1 + .01 + .001 etc....) Is actually convergent even if the sequence never stops growing.
Okay. I'll use the p-series. Thanks for your help!
No problem :)