Another convergent and divergent

[tnutty]

Homework Statement

Determine whether the series is convergent or divergent.
\sum n⁵ / (n⁶ + 1)

 

[lanedance]

hi tnutty - any ideas?
maybe a comparison test...

 

[tnutty]

Yes I was thinking of the comparison test, but that's 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.

Any ideas solving this by integral test?

 

[Mark44]

Yes I was thinking of the comparison test, but that's 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.
?

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 \sum 1/n.

 

[Mark44]

The comparison test does show divergence that's 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.

For the integral test however, since the numerator contains n^5 and the derivative of the denominator is 6n^5 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.

 

[lanedance]

as Mark mentioned out the comparison test points out that if

b_n &gt; a_n for all n>N then if a_n diverges so does b_n
so your pevious example doesn't work...

but to get this condition you could notice

\frac{n^5}{n^6 + 1} &gt; \frac{1}{2n} which is true \forall n &gt;1

or
\frac{n^5}{n^6 + 1} &gt; \frac{n^5}{n^6 + n^5} = \frac{1}{n+1} which is true \forall n &gt;1 and cleary diverges...

clearly