Another convergent and divergent

1. Mar 17, 2009

tnutty

1. The problem statement, all variables and given/known data
Determine whether the series is convergent or divergent.
$$\sum$$ n5 / (n6 + 1)

2. Mar 17, 2009

lanedance

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

3. Mar 17, 2009

tnutty

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.

Any ideas solving this by integral test?

4. Mar 17, 2009

rwisz

The comparison test does show divergence that's right.

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.

5. Mar 17, 2009

rwisz

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. And because I'm going to bed right now and you seem like you really do need some brushing up on your series, here's the work for it step-by-step!

$${\lim }\limits_{b \to \infty } \int_1^b \frac{n^5}{n^6+1}dn$$ Determine your u substitution.

$$u=n^6+1, du=6n^5$$ Rewrite the integral.

$${\lim }\limits_{b \to \infty } \frac{1}{6} \int_1^b \frac{du}{u}$$ Evaluate the integral.

$${\lim}\limits_{b \to \infty } ln(6b^6+1)-ln(6^6+1)$$ Substitute in for $$b$$

$$\infty - ln(6^6+1) = \infty$$

Infinity minus a number = Infinity.

Hence by the Integral Test the series diverges!

6. Mar 18, 2009

Staff: Mentor

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 $\sum 1/n$.

7. Mar 18, 2009

Staff: Mentor

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.

8. Mar 18, 2009

lanedance

as Mark mentioned out the comparison test points out that if

$$b_n > a_n$$ for all n>N then if an diverges so does bn
so your pevious example doesn't work...

but to get this condition you could notice

$$\frac{n^5}{n^6 + 1} > \frac{1}{2n}$$ which is true $$\forall n >1$$

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

clearly

9. Mar 18, 2009

rwisz

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? :)

No but in all seriousness I do apologize, for lack of a better phrase, my bad!