Could this solution be accepted?

[twoflower]

Hi, assume this sum:

$$\sum_{n = 1}^{\infty} \frac{ \sqrt{n^2 + n} - \sqrt[3]{n^3 + n}}{ \sqrt{n^3}}$$

So I chose this divergent sum:

$$b_{n} = \frac{1}{n}$$

to compare with my original sum. Then

$$\lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = \lim_{n \rightarrow \infty} \frac{ \sqrt{n^4 + n^3} - \sqrt[3]{n^6 + n^4}}{ \sqrt{n^3}} = \lim_{n \rightarrow \infty} \sqrt { \frac{ n^4 + n^3}{n^3}} - \lim_{n \rightarrow \infty} \sqrt[3]{ \frac{n^6 + n^4}{n^9}} = \infty - 0 = \infty$$

Thus my original sum diverges (is "larger" than the divergent sum 1/n).

Would this solution be accepted in test?

Thank you.

 

[WaR]

I'm not entirely sure what you did in the first step of you limit but

$$\lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}}$$ using your suggested $$a_{n}$$ and $$b_{n}$$ is equal to $$0$$ not $$\infty$$.

Your Limit Comparission Test would be inconclusive.

 

[twoflower]

Whooops I see it. Hm that's bad, I thought I had it :-(

 

[arildno]

Twoflower:
In order to handle this sum, you should first rewrite the expression for the terms as follows:
$$a_{n}=\frac{\sqrt{1+\frac{1}{n}}-\sqrt[3]{1+\frac{1}{n^{2}}}}{\sqrt{n}}$$

Your next step should be a clever use of Taylor series approximations to the two radicand expressions in your numerator, which both are of the following form (provided n is large:
$$(1+\epsilon)^{a},\epsilon<<1$$

You should be able to show that the Taylor series representations for fractional "a"'s are ULTIMATELY an alternating series with terms of decreasing magnitude.
Use that fact to bound your numerator; you should end up with the conclusion that your series is convergent.
good luck!

 

[shmoe]

You should be able to do a direct comparison test. The annoying thing is the roots in the numerator. Try bounding the polynomial $$n^2+n$$ by a polynomial that's a perfect square so you can remove this root. Aim for a perfect cube for the other one.

Remember the difference in sign, so if you are trying to bound the entire expression from above, bound $$n^2+n$$ from above and $$n^3+n$$ from below (vice versa if you want a lower bound-I'll leave it to you to determine whether you want an upper or a lower bound). Also remember that this bound only needs to hold for large n.

 

[twoflower]

Twoflower:
In order to handle this sum, you should first rewrite the expression for the terms as follows:
$$a_{n}=\frac{\sqrt{1+\frac{1}{n}}-\sqrt[3]{1+\frac{1}{n^{2}}}}{\sqrt{n}}$$

Your next step should be a clever use of Taylor series approximations to the two radicand expressions in your numerator, which both are of the following form (provided n is large:
$$(1+\epsilon)^{a},\epsilon<<1$$

You should be able to show that the Taylor series representations for fractional "a"'s are ULTIMATELY an alternating series with terms of decreasing magnitude.
Use that fact to bound your numerator; you should end up with the conclusion that your series is convergent.
good luck!

Thank you for the solution arildno, unfortunately I don't know Taylor series yet. Maybe I should learn them in advance when I see the amount of sums solvable with them...

 

[arildno]

Then you should use shmoe's suggestion.

 

[twoflower]

You should be able to do a direct comparison test. The annoying thing is the roots in the numerator. Try bounding the polynomial $$n^2+n$$ by a polynomial that's a perfect square so you can remove this root. Aim for a perfect cube for the other one.

Remember the difference in sign, so if you are trying to bound the entire expression from above, bound $$n^2+n$$ from above and $$n^3+n$$ from below (vice versa if you want a lower bound-I'll leave it to you to determine whether you want an upper or a lower bound). Also remember that this bound only needs to hold for large n.

Thank you shmoe, however I don't fully understand to the second part of your answer. Yes, I can multiply it with such expressions, so that I get this:

$$\sum_{n=1}^{ \infty} \frac{ n + 1 }{ \sqrt{ n^3 + n^2 } } - \frac{ n - 1 }{ \sqrt{ n^3 } \left( \sqrt[3]{ 1 - \frac{ 3n^2 - 2n + 1 }{ n^3 + n } } \right) }$$

Well, it seems yet more complicated to me...

 

[arildno]

In order to bound the terms along shmoe's lines, do as follows:
1) We first prove:
The 3rd root expression in the numerator is strictly less than the square root expression:
$$n(\sqrt{1+\frac{1}{n}}-\sqrt[3]{1+\frac{1}{n^{2}}})=n(\sqrt{1+\frac{1}{n}}-\sqrt{1+\frac{1}{n^{2}}}\frac{\sqrt[3]{1+\frac{1}{n^{2}}}}{\sqrt{1+\frac{1}{n^{2}}}})=n(\sqrt{1+\frac{1}{n}}-\frac{\sqrt{1+\frac{1}{n^{2}}}}{(1+\frac{1}{n^{2}})^{\frac{1}{6}}})\geq{n}(\sqrt{1+\frac{1}{n}}-\sqrt{1+\frac{1}{n^{2}}})\geq{0}$$

2) We bound therefore the first term by:
$$\sqrt{n^{2}+n}\leq\sqrt{(n+1)^{2}}=n+1$$
3) We bound the negative term as follows:
$$-\sqrt[3]{n^{3}+n}}\leq{-\sqrt[3]{n^{3}}}=-n$$
4)
Hence,
$$a_{n}\leq\frac{(n+1)+(-n)}{\sqrt{n^{3}}}=\frac{1}{n^{\frac{3}{2}}}$$

 

[twoflower]

Wow, now I'm clear about that. Your suggestions are always very "brain-freshing" for me, arildno :) Thank you.

 

[twoflower]

Yeah and one more question: why do we have to prove that the 3rd root expression in the numerator is strictly less than the square root expression? Wouldn't the two following bounds be sufficient?

 

[shmoe]

You need to show the terms are positive as well. Just knowing they are less than $$1/n^{3/2}$$ doesn't stop them from being very negative.

You can use arildno's method, or you can show a weaker $$\sqrt[3]{n^3 + n}-\sqrt{n^2 + n}<1$$ using bounds similar to our earlier ones. This will actually show $$|\sqrt{n^2 + n} - \sqrt[3]{n^3 + n}|<1$$ which will show your sum is absolutely convergent (which you'd know anyway if you had proven it was positive).

 

[twoflower]

Ok I have it now, thank you shmoe!