moment of iid sums Let Let $X_1,X_2,X_3,X_4$ be a sequence of independent, identically distributed random variables
with:
$$ E(X) = 0 $$
$$ E(X^2)=1$$
$$E(X^3) = 1$$
$$E(X^4) = 6 $$
Let:
$$S_1 =X_1$$
$$S_2 =X_1+X_2$$
$$S_3 = X_1+X_2+X_3$$
and so on.
Show that 
$$\sum_{r=1}^{n} \frac{E(S_r^4)}{r^2(r+1)^2} = \frac{3n}{n+1}$$
Using central moments of iids
I have found the 4th central moment of $S_n$ to be
$$E(S_n^4) = n(6-3n^2)$$
putting this into the sum i have
$$\sum_{r=1}^{n}\frac{r(6-3r^2)}{r^2(r+1)^2}$$
not really sure where to go from here in order to get the expression needed
 A: For a start, condider $S_r^4 = (X_1 + X_2 + \cdots + X_r)^4$. Expanding this, we will get terms like, for example, $X_1 X_2^3$, where the expectation value will be $E[X_1 X_2^3] = E[X] E[X^3] $  since the variables are iid. In this case, the expectation value is zero since  $E[X] =0$. 
Generalizing this example, in the expansion of $S_r^4$we need only to look for terms with $X_p^4$  and  $X_p^2 X_q^2$ ($p \ne q)$ since in all other terms, which are products of 4 variables, one variable $X_p$ will occur with power $1$ and hence the expectation value of this term will be zero.
Now in the expansion of $S_r^4$  there are $r$ terms  with $X_p^4$  and $6 \binom{r}{2}$ terms  with  $X_p^2 X_q^2$ ($p \ne q)$. So we can now write
$$
\sum_{r=1}^{n} \frac{E[S_r^4]}{r^2(r+1)^2} = \sum_{r=1}^{n} \frac{r E[X^4] + 6 \binom{r}{2} (E[X^2] )^2}{r^2(r+1)^2} 
$$
With the given values, this is 
$$
 \sum_{r=1}^{n} \frac{6 r + 6 \binom{r}{2} }{r^2(r+1)^2} = 3 \sum_{r=1}^{n} \frac{2 r + r (r-1)}{r^2(r+1)^2} = 3 \sum_{r=1}^{n} \frac{1}{r(r+1)} 
$$
and the last expression can be easily summed to be $3 \frac{n}{n+1}$. This proves the claim. $\qquad \Box$
