Find: $\zeta \left( 3,1,1,1 \right)$ While solving a summation, I came across this: 

$$\zeta \left( 3,1,1,1 \right)=? $$

I'm new to multiple zeta values. That's why I couldn't find this. So my question is does a closed form exist for this. If yes, what is it? I have no idea to evaluate this. Please help.
 A: It is known that every multiple zeta value can be expressed as a sum of MZVs of the same weight, where the arguments are 2's and 3's only.  (Here weight means the sum of the arguments, so $ 1 + 1 + 1 + 3 = 6 $ in this case.)  This is called the Hoffman basis,  and this result is proven by Francis Brown in the paper "Mixed Tate motives over $\mathbb{Z}$".
The upshot is that you can express every MZV of weight 6 as a combination of $ \zeta(2,2,2) $ and $ \zeta(3,3) $.  Then you can use that $ \zeta(2,2,2) = \frac{1}{7!} \pi^6 $ (see below), and the 'stuffle product' relation $ \zeta(3)^2 = \zeta(6) + 2\zeta(3,3) $.  This allows you to get some combination of $ \pi^6 $ (or $ \zeta(6) $, or $ \zeta(2)^3 $ if you prefer) and $ \zeta(3) $ instead.
Specifically you find
$$
\zeta(3,1,1,1) = -\frac{1}{2} \zeta(3)^2 + \frac{3}{4} \zeta(6) \, .
$$
You can then use $ \zeta(6) = \frac{1}{945} \pi^6 $, if you want to write the result in terms of $ \pi $ instead.
You can use the tool EZ-Face to numerically confirm this, and this is perfectly sufficient to claim an identity at such a low weight.  This command will find any linear dependencies between the inputs:

lindep([z(3)*z(3), z(6), z(3,1,1,1)])

The output is:

[2., -3., 4.]

This indicates that $ 2\zeta(3)^2 - 3\zeta(6) + 4 \zeta(3,1,1,1) = 0 $, which can be rearranged to the evaluation above.

Edit: The result that $\zeta(2,2,2) = \frac{1}{7!} \pi^6 $, or more generally
$$
    \zeta(\underbrace{2, \ldots, 2}_{\text{\( k \) times}}) = \frac{1}{(2k+1)!} \pi^{2k} \,
$$
can be found in Borwein, Bradley and Broadhurst's paper "Evaluations of $k$-fold Euler/Zagier sums".  It follows by slightly simplifying equation (36).
You can find a proof for the specific case $\zeta(2,2,2)$ directly with the `stuffle product'.  We have
$$
  \zeta(2)\zeta(2) \overset{\text{stuffle}}{=} \zeta(4) + 2 \zeta(2,2)
$$
Multiply both sides by $ \zeta(2) $ and use
$$
  \zeta(2,2)\zeta(2) \overset{\text{stuffle}}{=} 3 \zeta(2,2,2) + \zeta(4,2) + \zeta(2,4) \,
$$
to write
$$
 \zeta(2)^3 = \zeta(2)\zeta(4) + 6\zeta(2,2,2) + 2\zeta(4,2) + 2\zeta(2,4) \, .
$$
Finally use
$$
  \zeta(4)\zeta(2) \overset{\text{stuffle}}{=} \zeta(6) + \zeta(4,2) + \zeta(2,4)
$$
to get rid of the $ \zeta(4,2) $ and $ \zeta(2,4) $.
Now you have
$$
  \zeta(2)^3 = 3\zeta(2)\zeta(4) + 6 \zeta(2,2,2) - 2 \zeta(6) \, .
$$
You can solve this for $ \zeta(2,2,2) $ and use Euler's evaluation of $ \zeta(\text{even}) $ to evaluate the result.
A: Applying the sum formula
$$
\sum_{\sum a_i=n,\, a_i\ge 0}\zeta(a_1+2,a_2+1,...,a_r+1)=\zeta(n+r+1).  
$$
for $r=4$ and $n=1$ we obtain 
$$ 
\zeta(3,1,1,1)+\zeta(2,2,1,1)+\zeta(2,1,2,1)+\zeta(2,1,1,2) =\zeta(6).
$$
I found in the literature that $\zeta(2,1,2,1)=\zeta(3,3)=\frac{\zeta(3)^2-\zeta(6)}{2}$, and similarly 
$$
\zeta(2,2,1,1)=\zeta(3)^2-\frac{4}{3}\zeta(6),\; \zeta(2,1,1,2)=\frac{25\zeta(6)-12\zeta(3)^2}{12},
$$
so that 
$$
\zeta(3,1,1,1)=\frac{3\zeta(6)-2\zeta(3)^2}{4}.
$$
