# Divergent Series - How to test it

## Homework Statement

Ʃ (1+n+n2)/(√1+n2+n6)) n=1 to infinity

## Homework Equations

Comparison Test / P Series
0≤an≤bn

## The Attempt at a Solution

Hello, I simplified the problem to

Ʃ (1+n+n2)/(1+n2+n6)1/2

Is it incorrect of me to say immediately right here that because the power of the denominator p<1 it is divergent?

Or if I take a different approach use comparison test and say 0≤an≤1/(1+n2+n6)1/2 can I say it is divergent because bn is divergent due to p-series?

WA says it is divergent due to comparison test but how would I get something like say 1/n^6 out of that square root to use as bn?

Simon Bridge
Homework Helper
I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.

Dick
Homework Helper
I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.

True, but 1/n is on the boundary between divergence and convergence for power series. You need an actual comparison to make. Being asymptotically 'like' 1/n is not good enough. Choose a series with a smaller denominator and a larger denominator and show it diverges. I'm a little worried that what Saterial has said so far makes little sense.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Ʃ (1+n+n2)/(√1+n2+n6)) n=1 to infinity

## Homework Equations

Comparison Test / P Series
0≤an≤bn

## The Attempt at a Solution

Hello, I simplified the problem to

Ʃ (1+n+n2)/(1+n2+n6)1/2

Is it incorrect of me to say immediately right here that because the power of the denominator p<1 it is divergent?

Yes, that would be incorrect.

Or if I take a different approach use comparison test and say 0≤an≤1/(1+n2+n6)1/2 can I say it is divergent because bn is divergent due to p-series?

No. It isn't a p series

WA says it is divergent due to comparison test but how would I get something like say 1/n^6 out of that square root to use as bn?

Look up the "general comparison test" or it may be called the "limit comparison test" in your text.

LCKurtz
Homework Helper
Gold Member
I would have said it was divergent because it behaves as 1/n for large n... and we know the sum of 1/n diverges from a previous proof.

That would be a correct intuitive argument, but you need the limit comparison test to prove it.

Dick
Homework Helper
That would be a correct intuitive argument, but you need the limit comparison test to prove it.

A direct comparison test will do it. And it's simpler.

LCKurtz
Homework Helper
Gold Member
A direct comparison test will do it. And it's simpler.

Probably so and point taken. I haven't actually worked on it. But I do think it is good for students to recognize situations that are a natural setting for the limit comparison test.

Dick
Homework Helper
Probably so and point taken. I haven't actually worked on it. But I do think it is good for students to recognize situations that are a natural setting for the limit comparison test.

I'll repeat what I said before. Make the denominator smaller and the numerator larger in a way that the resulting series still diverges and you've got it. Limit comparison test may not be in Saterial's repetoire. And it's not needed. It's really simple once you've seen it.

Last edited:
Curious3141
Homework Helper
I'll repeat what I said before. Make the denominator smaller and the numerator larger in a way that the resulting series still diverges and you've got it. Limit comparison test may not be in Saterial's repetoire. And it's not needed. It's really simple once you've seen it.

Making the denominator smaller and the numerator larger (and hence every term larger) would not help you prove divergence.

Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?

Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?

Yes, the divergence follows rather trivially when one does that because the series collapses into the harmonic series.

(1+n+n2)/(√1+n2+n6)) > (1)/(√1+n2+n6)) > (1)/(√n2+n2+n6)) = (1)/(√2n2+n6)) = (1)/(√n2(2+n4))) = (1)/(n√(2+n4))) > 1/n

Dick
Homework Helper
Making the denominator smaller and the numerator larger (and hence every term larger) would not help you prove divergence.

Maybe you meant make the denominator larger and the numerator smaller, and prove that the resulting series still diverges?

Oh yeah. That's what I meant. Denominator is the thing on the BOTTOM. Duh. Thanks!

Dick
Homework Helper
(1+n+n2)/(√1+n2+n6)) > (1)/(√1+n2+n6)) > (1)/(√n2+n2+n6)) = (1)/(√2n2+n6)) = (1)/(√n2(2+n4))) = (1)/(n√(2+n4))) > 1/n

Sorry. (1)/(n√(2+n4))) is less than 1/n. Not greater.

Dick
Homework Helper
How about just noting that the original series is greater than n^2/sqrt(n^6+n^6+n^6)?

Curious3141
Homework Helper
How about just noting that the original series is greater than n^2/sqrt(n^6+n^6+n^6)?

Yes, that would work. Much simpler, too.

I just showed the first thing I thought of.

uart
Or as another alternative noting that it's greater than $\frac{n^2}{n^3 + 1}$, which in turn is greater than $\frac{1}{n+1}$. We can then compare it directly to the harmonic series starting at n=2.