Concrete Mathematics: How do we figure out the constrains of summations when using multiplication by summation factor method? In chapter 2.2 of Concrete Mathematics, the authors introduced the usage of summation factor to convert recurrence to summation. The idea is to multiply $s_n$ on both sides of the recurrence relation of the form $a_n T_n = b_n T_{n-1} + c_n$ to obtain $s_n a_n T_n = s_n b_n T_{n-1} + s_n c_n$. If we select $s_n b_n = s_{n-1} a_{n-1}$, and let $S_n = s_n a_n T_n$, then we get $S_n = S_{n-1} + s_n c_n$, which can be expressed in terms of a sum easily.
Here's my question. When they are solving for the quick sort recurrence $C_0 = C_1 = 0$ and $C_n = n + 1 + \frac{2}{n} \sum_{k=0}^{n-1} C_k$, the final closed form formula have a restriction of $n > 1$. Why?
More general, how can we figure out the restriction of $n$ while using such technique?
EDIT
In order to solve for $C_n = n + 1 + \frac{2}{n} \sum_{k=0}^{n-1} C_k$, first multiply both sides by $n$ to obtain $nC_n = n^2 + n + 2 \sum_{k=0}^{n-1} C_k$. If you replace $n$ by $n-1$, we get $(n-1)C_{n-1} = (n-1)^2 + (n-1) + 2 \sum_{k=0}^{n-2} C_k$. Take note this is only valid if $n>2$. Subtract both equations, we get $nC_n =(n+1)C_{n-1}+2n$ for $n>2$, and $C_0 = C_1 = 0$ and $C_2 = 3$. From this point, the recurrence will only be valid for $n>2$, but the final solution is valid for $n>1$. Why?

The summation factor is $\frac2{n(n+1)}$, so we get $S_n=\frac2{n+1}C_n$ and $S_n=S_{n-1}+\frac4{n+1}$ for $n\ge 3$. $C_2=3$, so $S_2=\frac23C_2=2$, and
$$S_n=2+\sum_{k=3}^n\frac4{k+1}=2+4(H_{n+1}-H_3)=4H_{n+1}-\frac{16}3$$
for $n\ge 3$. Thus,
$$\begin{align*}
C_n&=2(n+1)H_{n+1}-\frac83(n+1)\\
&=2(n+1)\left(H_n+\frac1{n+1}\right)-\frac83n-\frac83\\
&=2(n+1)H_n-\frac83n-\frac23\tag{1}
\end{align*}$$
for $n\ge 3$.
The derivation is valid only for $n\ge 3$, but the expression $(1)$ might by good fortune be valid for smaller values of $n$, and one should always test. Here we find that for $n=2$ it evaluates to
$$2\cdot3\cdot\frac32-\frac{16}3-\frac23=9-6=3\;,$$
which actually is $C_2$. However, at $n=1$ it evaluates to
$$2\cdot2\cdot1-\frac83-\frac23=4-\frac{10}3\ne 0=C_1\;,$$
so the closed form 
$$C_n=2(n+1)H_n-\frac83n-\frac23$$
is valid precisely for $n\ge 2$. Its validity for $n=2$ doesn’t follow from the derivation: it follows from an independent verification after the fact.
