sequences and series - Riemann $zeta(3)$ convergence with Cauchy

I'm an undergraduate freshman math student, and we were asked to prove that the sequence $a{_n} =\sum_{k=1}^{n} \frac{1}{k^3}$ converges (obviously, we weren't asked to calculate its limit.) Our teacher hinted to prove that it's a Cauchy sequence. We don't know much, only Cauchy, several sentences about sequences and limits and monotonic sequences and such (basic first few months of undergraduate freshman). I'm stuck. any hints / ideas?

Here's my attempt:

Let $\varepsilon > 0$. We need to find N, such that for all $m > n > N$, $a_{m}-a_{n} < \varepsilon$. $a_{m}-a_{n} = \sum_{k = n+1}^{m} \frac{1}{k^3}$.

$\sum_{k = n+1}^{m} \frac{1}{k^3} < \frac{m-n}{(n+1)^3}$.

But this leads nowhere.

Note: We don't have to prove it by Cauchy, any solution (from the little we have learnt) will do.

Answer

For $k\geq 2$ we have $k^2\geq k+1$

and

$$\frac{1}{k^3}\leq \frac{1}{k(k+1)}$$

but

$$\sum_{k=2}^n\frac{1}{k(k+1)}=\sum_{k=2}^n (\frac{1}{k}-\frac{1}{k+1})$$

$$=\frac{1}{2}-\frac{1}{n+1}\leq \frac{1}{2}$$

thus the sequence of partial sums

$S_n=\sum_{k=2}^n\frac{1}{k^3}$ is increasing and bounded, and therefore convergent.