Series to compare to for comparison test

[ToNoAvail27]

Homework Statement

Does \sum_{n=1}^{\infty}a_n where a_n = \frac{(n+1)^{1/3}-n^{1/3}}{n} converge or diverge?

Homework Equations

The Attempt at a Solution

The ratio test is inconclusive, as is the root test. The limit is equal to 0, but that doesn't say much. I've tried to find another series to compare it too, but with the cube roots, I'm having a bit of trouble. I have, but they've all been bigger and have diverged, which again doesn't help. Usually, I can guess one by seeing if one term takes over the other over time, but this one is stumping me. Could I perhaps get a hint as to what to consider in order to find a suitable series? I just need a small push in the right direction. Thanks for your time

 

[PeroK]

There's always the Binomial Theorem!

 

[Dick]

Homework Statement

Does \sum_{n=1}^{\infty}a_n where a_n = \frac{(n+1)^{1/3}-n^{1/3}}{n} converge or diverge?

Homework Equations

The Attempt at a Solution

The ratio test is inconclusive, as is the root test. The limit is equal to 0, but that doesn't say much. I've tried to find another series to compare it too, but with the cube roots, I'm having a bit of trouble. I have, but they've all been bigger and have diverged, which again doesn't help. Usually, I can guess one by seeing if one term takes over the other over time, but this one is stumping me. Could I perhaps get a hint as to what to consider in order to find a suitable series? I just need a small push in the right direction. Thanks for your time

If they were square roots I'd multiply numerator and denominator by the conjugate to 'rationalize' it. Can you think of a way to apply $a^3-b^3=(a-b)(a^2+ab+b^2)$ to accomplish the same thing with cube roots?

 

[ToNoAvail27]

If they were square roots I'd multiply numerator and denominator by the conjugate to 'rationalize' it. Can you think of a way to apply $a^3-b^3=(a-b)(a^2+ab+b^2)$ to accomplish the same thing with cube roots?

But of course! =D

\frac{(n+1)^{1/3}-n^{1/3}}{n} = \frac{(n+1)^{1/3}-n^{1/3}}{n} * \frac{(n+1)^{2}+n(n+1)+n^{2}}{(n+1)^{2}+n(n+1)+n^{2}} = \frac{n+1-n}{n[(n+1)^{2}+n(n+1)+n^{2}]} = \frac{1}{3n^{3}+3n^{2}+n}. Since n\ge 1, 3n^{3} < 3n^{3}+3n^{2}+n \Rightarrow \frac{1}{3n^{3}+3n^{2}+n} < \frac{1}{3n^{3}}, the later of which converges since its exponent is greater than 1! Thanks!

 

[PeroK]

There's something not quite right there. There should be fractional powers, surely?

 

[Dick]

But of course! =D

\frac{(n+1)^{1/3}-n^{1/3}}{n} = \frac{(n+1)^{1/3}-n^{1/3}}{n} * \frac{(n+1)^{2}+n(n+1)+n^{2}}{(n+1)^{2}+n(n+1)+n^{2}} = \frac{n+1-n}{n[(n+1)^{2}+n(n+1)+n^{2}]} = \frac{1}{3n^{3}+3n^{2}+n}. Since n\ge 1, 3n^{3} < 3n^{3}+3n^{2}+n \Rightarrow \frac{1}{3n^{3}+3n^{2}+n} < \frac{1}{3n^{3}}, the later of which converges since its exponent is greater than 1! Thanks!

Your 'conjugate' part is wrong. For example $((n+1)^{1/3})^2$ isn't equal to $(n+1)^2$.

 

[ToNoAvail27]

Oops
I guess in my "excitement" I neglected the powers.
So we ACTUALLY obtain \frac{1}{n[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3}]}, and now n^{2/3} < (n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3} so that \frac{1}{n[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3}]} < \frac{1}{n^{\frac{5}{3}}}, which converges by the p-test. Is this now correct, or am I still going too fast and making mistakes? Thanks everyone for the comments so far.

 

[Dick]

Oops
I guess in my "excitement" I neglected the powers.
So we ACTUALLY obtain \frac{1}{n[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3}]}, and now n^{2/3} < (n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3} so that \frac{1}{n[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3}]} < \frac{1}{n^{\frac{5}{3}}}, which converges by the p-test. Is this now correct, or am I still going too fast and making mistakes? Thanks everyone for the comments so far.

Yes, that looks much better.