Prove that the sum is less than $6/5$. How to prove that $$\sum_{n=1}^{15} \frac{1}{n^3}\lt\frac 65$$ 

I tried to compare this sum to the infinite sum, but Apery's constant is just above $1.2$ so this approach doesn't work.

Then I typed this into wolfy and the sum seems as if it is just under $6/5$. 

However, I was wondering if there was a better way of proving this result rather than summing all the terms.
You can estimate the remainder of the series by comparing it to the integral.

$\displaystyle \sum\limits_{n=N}^{\infty}\frac 1{n^3}\ge\int_{N}^{\infty}\frac{dt}{t^3}=\bigg[\frac{-1}{2t^2}\bigg]_{N}^{+\infty}=\frac{1}{2N^2}$

$\displaystyle \sum\limits_{n=1}^{15}\frac 1{n^3}\le\zeta(3)-\frac 1{512}\simeq 1.2001... &gt; \frac 65$

$\displaystyle \sum\limits_{n=1}^{14}\frac 1{n^3}\le\zeta(3)-\frac 1{450}\simeq 1.1998... &lt; \frac 65$

Unfortunately the integral comparison cannot tell with certitude if the proper $N$ is $15$, so unless there are other ideas, I think the effective computation is required.
$\sum_{N+1}^{15} (1/x^3) $ &lt; $\int_N^{15} (1/x^3) dx $  Try N=1 ,if it doesn't work,try N=2 etc this comes from the integral test . In fact you should   increase the upper limit of the integral from 15 to  +infinity  to make it easier .I believe N=3 might work