Proving that $(\sum_{j=1}^{n} X_{j}) / b_{n} \overset {P}{\to} C$ implies $b_{n}\sim b_{n+1}$ I would like some help proving the following result. Thanks for any help in advance.

Let $(X_{n})_{n=1}^\infty$ be i.i.d. random variables, $C$ some nonzero constant, and let $(b_n)$ a sequence of positive reals such that $\lim_n b_n = \infty$. Prove that if $(\sum_{j=1}^{n} X_{j}) / b_{n} \overset {P}{\to} C$, then $b_{n}\sim b_{n+1}$.

 A: A standard result on the convergence of vectors yields the convergence in probability of $\left(\frac{1}{b_n} \sum_{j=1}^n X_j,\frac{1}{b_{n+1}} \sum_{j=1}^{n+1} X_j\right)$ to  $(C,C)$ . The function $(x,y)\mapsto \frac yx$ has discontinuity set $D:=\{(0,y), y\in\mathbb R\}$ and $P((C,C)\in D)=0$, so the continuous mapping theorem applies and $$\frac{b_n}{b_{n+1}}\left( 1+ \frac{X_{n+1}}{\sum_{j=1}^n X_j}\right)\xrightarrow[n\to \infty]{P} 1$$

Let us prove that $\displaystyle \frac{X_{n+1}}{\sum_{j=1}^n X_j}\xrightarrow[n\to \infty]{P} 0$. Let $\epsilon >0$ and note that
$$ \left(|X_{n+1}|\geq \epsilon \left|\sum_{j=1}^n X_j\right|\right)
= A \cup B
$$
with $A=\left(\frac{|X_{n+1}|}{b_n}\geq \epsilon \frac{\left|\sum_{j=1}^n X_j\right|}{b_n}\right)\cap \left(\left|\frac{\sum_{j=1}^n X_j}{b_n}-c\right| \geq \frac{|c|}2\right) $ and
$B=\left(\frac{|X_{n+1}|}{b_n}\geq \epsilon \frac{\left|\sum_{j=1}^n X_j\right|}{b_n}\right)\cap \left(\left|\frac{\sum_{j=1}^n X_j}{b_n}-c\right| < \frac{|c|}2\right) $.
With the inclusions $\displaystyle A\subset \left(\left|\frac{\sum_{j=1}^n X_j}{b_n}-c\right| \geq \frac{|c|}2\right)$ and $\displaystyle B\subset \left(\frac{|X_{n+1}|}{b_n}\geq \frac{\epsilon |c|}2 \right)$, one gets the bound
$$P\left(|X_{n+1}|\geq \epsilon \left|\sum_{j=1}^n X_j\right|\right)\leq \underbrace{P\left(\left|\frac{\sum_{j=1}^n X_j}{b_n}-c\right| \geq \frac{|c|}2\right)}_{\to 0} + P\left(\frac{|X_{n+1}|}{b_n}\geq \frac{\epsilon |c|}2 \right) $$
Since the $X_i$ are identically distributed and $b_n\to \infty$, $$P\left(\frac{|X_{n+1}|}{b_n}\geq \frac{\epsilon |c|}2 \right) = P\left(|X_1|\geq b_n\frac{\epsilon |c|}2\right) \xrightarrow[n\to \infty]{}0$$
Hence $\displaystyle \frac{X_{n+1}}{\sum_{j=1}^n X_j}\xrightarrow[n\to \infty]{P} 0$.

By the continuous mapping theorem, $\displaystyle \frac{1}{1+\frac{X_{n+1}}{\sum_{j=1}^n X_j}} \xrightarrow[n\to \infty]{P} 1$, hence $$\frac{b_n}{b_{n+1}}\left( 1+ \frac{X_{n+1}}{\sum_{j=1}^n X_j}\right) \frac{1}{1+\frac{X_{n+1}}{\sum_{j=1}^n X_j}}\xrightarrow[n\to \infty]{P} 1$$that is $\displaystyle \frac{b_n}{b_{n+1}} \xrightarrow[n\to \infty]{P} 1$. As $\displaystyle \frac{b_n}{b_{n+1}}$ is a deterministic sequence, this easily implies $$\frac{b_n}{b_{n+1}}\xrightarrow[n\to \infty]{} 1$$ 
Note that the independence assumption on the $X_i$ was not used.
