No, you can't conclude from this comparison that the series diverges. For the comparison test to show that a series diverges, the terms have to be larger than those of a divergent series. Here you show that they are smaller than those of [itex]\sum 1/n[/itex].Yes I was thinking of the comparison test, but thats next chapter. in this chpt, its all about integral test. i am not sure how to solve this with integral test, but can you check out the comparison test that follows ?
Comparison test ;
n^5 / (n^6+1) <= n^5 / n^6 = 1/n and from definition we know that 1/(n^p)
converges if n > 1 and diverges if n< 1. So in this case it diverges since n = 1.
Sorry to burst your bubble, but no it does not. Take a look at the comparison test and what it says about divergent series and what it says about convergent series. They are different.The comparison test does show divergence that's right.
For the integral test however, since the numerator contains [tex]n^5[/tex] and the derivative of the denominator is [tex]6n^5[/tex] then you should be able to tell that u-substitution will work like a charm here...
Hint: du/u = ln u.
And for the setup of the improper integral, try looking at the previous thread where I helped you, at the bottom of my last post.
Ah good catch! I was not paying close enough attention, I apologize. But I did prove divergence by the Integral test... so I get some slack right? :)Sorry to burst your bubble, but no it does not. Take a look at the comparison test and what it says about divergent series and what it says about convergent series. They are different.