# Series to compare to for comparison test

1. Nov 28, 2013

### ToNoAvail27

1. The problem statement, all variables and given/known data
Does $\sum_{n=1}^{\infty}a_n$ where $a_n = \frac{(n+1)^{1/3}-n^{1/3}}{n}$ converge or diverge?

2. Relevant equations

3. 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

2. Nov 28, 2013

### PeroK

There's always the Binomial Theorem!

3. Nov 28, 2013

### Dick

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?

4. Nov 29, 2013

### ToNoAvail27

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!

5. Nov 29, 2013

### PeroK

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

6. Nov 29, 2013

### Dick

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

7. Nov 29, 2013

### 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.

8. Nov 29, 2013

### Dick

Yes, that looks much better.