Proving a sequence is convergent to a specific value 
Suppose $\{a_n\}$ is a strictly increasing sequence in $\mathbb{R}$ and also 
  $\{x_n\}$ is a sequence which is convergent to $x$.
if $$\lim_{n\to\infty}\sum_{i=1}^{n}a_i=\infty$$
  Then prove $\{\dfrac{\sum_{i=1}^n a_ix_i}{\sum_{i=1}^n a_i}\}\to x$

I tried to prove it with some inequalities and also using basic epsilon-delta form, but neither of them was helpful.
 A: We have an $N$ such that for $n\ge N$, $|x_n - x| < \epsilon/2$ and an $N'$ such that for $n\ge N'$, $$\sum_{i=1}^n a_i > \frac{2 \left|\sum_{i=1}^N a_i(x_i - x)\right|}{\epsilon}$$ Let $N'' = \max\{N,N'\}$. Thus, for $n\ge N''$,
$$\left| \frac{\sum_{i=1}^n a_i(x_i - x)}{\sum_{i=1}^n a_i} \right| < \epsilon$$
Which gives the result.
A: Let $A_n = \sum_{k=1}^n a_kx_k$ and $B_n = \sum_{k=1}^n a_k$. 
Then $B_n \to \infty$ and
$$\lim_{n \to \infty} \frac{A_{n+1} - A_n}{B_{n+1} - B_n} = \lim_{n \to \infty}x_{n+1} = x.$$
Simply apply the Stolz-Cesaro theorem or follow this proof.
For $n > N$ we have
$$\left|\frac{A_{n+1} - A_n}{B_{n+1} - B_n} - x\right| < \frac{\epsilon}{2},$$
and
$$\left|\frac{A_{N+1} - A_N}{B_{N+1} - B_N} - x\right| < \frac{\epsilon}{2}  \implies \left|\frac{A_{n} - A_N}{B_{n} - B_N} - x\right| < \frac{\epsilon}{2}.$$
Thus,
$$\frac{A_n}{B_n} - x = \frac{A_{N} - xB_N}{B_n} +\left(1 - \frac{B_N}{B_n} \right)\left(\frac{A_{n} - A_N}{B_{n} - B_N} - x \right),$$
and
$$\left|\frac{A_n}{B_n} - x\right| \leqslant \frac{|A_{N} - xB_N|}{B_n} +\left|1 - \frac{B_N}{B_n} \right| \, \left|\frac{A_{n} - A_N}{B_{n} - B_N} - x \right| \\  < \frac{|A_{N} - xB_N|}{B_n} + \frac{\epsilon}{2}.$$
Since $B_n \to + \infty$, the first term on the RHS of the inequality chain can be made smaller than $\epsilon/2$ for sufficiently large $n$.  
Hence, for all $\epsilon > 0$ we have for sufficiently large $n$
$$\left|\frac{A_n}{B_n} - x\right| < \epsilon $$
A: Let $\epsilon>0$ and choose $N$ such that $a_n >0$ and $|x-x_n| < \epsilon$ for all $n \ge N$.
Let $c_n = {\sum_{k=1}^n a_k x_k \over  \sum_{k=1}^n a_k } $.
If $n > N$ then
${\sum_{k <N} a_k x_k + (x-\epsilon)\sum_{k=N}^n a_k \over \sum_{k <N} a_k + \sum_{k=N}^n a_k } < c_n ={\sum_{k <N} a_k x_k + \sum_{k=N}^n a_k x_k \over \sum_{k <N} a_k + \sum_{k=N}^n a_k } < {\sum_{k <N} a_k x_k + (x+\epsilon)\sum_{k=N}^n a_k \over \sum_{k <N} a_k + \sum_{k=N}^n a_k }$.
If we let $n \to \infty$ we see that the left- and right-most terms
converge to $x-\epsilon, x+\epsilon$ respectively and so
$x-\epsilon \le \liminf_n c_n \le \limsup_n c_n \le x+\epsilon$.
Since $\epsilon$ was arbitrary, we see that
$x = \liminf_n c_n = \limsup_n c_n = x$ and so $\lim_n c_n = x$.
