Prove that the divergence of a sequence implies the divergence of another sequence For positive, monotone $c_n$'s, $$\frac{x_0+x_1+\cdots + \ x_n}{n+1} \to \xi$$ 
implies
$$\frac{c_0x_0 + c_1x_1 + \cdots + c_nx_n}{c_0 + c_1 + \cdots + c_n} \to \xi\text,$$
provided $\left (\frac{nc_n}{C_n} \right)$ is bounded and $C_n \to +\infty$ where $C_n = c_0+c_1+ \cdots + c_n$.
I can prove the case for $\xi$ finite. Does this also hold for $\xi$ infinite? I can't adapt the previous proof to this case. The book doesn't assume $\xi$ is finite, so I decided to look at this case too. What do you suggest? Note that it suffices to show that it holds for $+\infty$
 A: So I think I figured this out for monotone decreasing $c_n$: It's easy to see that $$\sum_{j=0}^{j=n}c_jx_j  = c_nS_n+\sum_{j=0}^{j=n-1}(c_j-c_{j+1})S_j$$ where $S_n = x_0+\cdots +x_n$.
For any $G > 0$, we are given that $S_n > (n+1)G$ for $n$ greater than a suitable $n_0$.
Thus we may write, putting the first sum for brevity equal to $N$:$$\frac{N}{C_n} \ge \frac{Gc_n(n+1)}{C_n} + \frac{\sum_{j=0}^{n_0}(c_j-c_{j+1})S_j}{C_n}+\frac{\sum_{j=n_0+1}^{n-1}(c_j-c_{j+1})G(j+1)}{C_n}$$since for $j > n_0$ we have $S_j > (j+1)G$. Now the second sum as $n$ grows goes to $0$ since $n_0$ is fixed. Let's say that its minimum value is $-\epsilon$ from a certain $n$ onwards. For the second sum, we may employ the first identity above, with $x_j = 1$ and obtain:$$\frac{N}{C_n} \ge \frac{Gc_n(n+1)}{C_n} - \epsilon \ + G\frac{(C_n-c_n(n+1))-(C_{n_0+1}-c_{n_0+1}(n_0+2))}{C_n}$$We again see that since $n_0$ is fixed and $C_n \to +\infty$ the last fraction can be broken into:$$\frac{N}{C_n} \ge G-\epsilon \ + G(-\epsilon')$$   for sufficiently large $n$ with $\epsilon$ and $\epsilon'$ arbitrary. This should conclude the proof. We can again start from $G' > 0$ and choose $G$ such that the last expression is greater than $G'$.
