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.
 A: You can estimate the remainder of the series by comparing it to the integral.
$\displaystyle \sum\limits_{n=N}^{\infty}\frac 1{n^3}\ge\int_{N}^{\infty}\frac{dt}{t^3}=\bigg[\frac{-1}{2t^2}\bigg]_{N}^{+\infty}=\frac{1}{2N^2}$
$\displaystyle \sum\limits_{n=1}^{15}\frac 1{n^3}\le\zeta(3)-\frac 1{512}\simeq 1.2001... > \frac 65$
$\displaystyle \sum\limits_{n=1}^{14}\frac 1{n^3}\le\zeta(3)-\frac 1{450}\simeq 1.1998... < \frac 65$
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.
A: This is just 
a numerical coincidence,
as far as I am concerned.
As others have noted,
according to Wolfy,
the sum to 15 terms is
$$\dfrac{56154295334575853}{46796108014656000}
=1.199977898097614804322644896619336727642969815345853855758...
$$
and the sum to 16 terms is
$$\dfrac{449325761325072949}{374368864117248000}
=1.200222038722614804322644896619336727642969815345853855758...
$$
A: This is probably not an answer to the question.
Consider $$\sum_{n=1}^{p} \frac{1}{n^3}=H_p^{(3)}$$ where appear generalized harmonic numbers and use the corresponding asymptotics
$$H_p^{(3)}=\zeta (3)-\frac{1}{2 p^2}+\frac{1}{2 p^3}-\frac{1}{4 p^4}+\frac{1}{12
   p^6}-\frac{1}{12 p^8}+O\left(\frac{1}{p^{10}}\right)$$ where $\zeta (3)\approx 1.20205690316$. Using this value for order $O\left(\frac{1}{p^{k}}\right)$ and computing for $p=15$, we should get
$$\left(
\begin{array}{cc}
 k & H_{15}^{(3)}\\
0,1 & 1.20205690316 \\
 2 & 1.19983468094 \\
 3 & 1.19998282909 \\
 4,5 & 1.19997789081 \\
 6,7 & 1.19997789813 \\
 8,9 & 1.19997789810 
 \end{array}
\right)$$
A: $\sum_{N+1}^{15} (1/x^3) $ < $\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
