Cesàro's Lemma - precise definition of limit, indices

Question

Probability with Martingales:

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1. I'm a little confused about the precise langauge.

I guess we have that

$$\forall M > 0, \exists N_b > 0 \ \text{s.t.} \ b_n > M \ \text{whenever} \ n > N_b \tag{1}$$

$$\forall \epsilon > 0, \exists N_{nv} > 0 \ \text{s.t.} \ |v_n - v_{\infty}| < \epsilon \ \text{whenever} \ n > N_{nv} \tag{2}$$

$$\forall \epsilon > 0, \exists N_{kv} > 0 \ \text{s.t.} \ |v_k - v_{\infty}| < \epsilon \ \text{whenever} \ n \ge k > N_{kv} \tag{3}$$

$$\forall \epsilon > 0, \exists N_{kv} > 0 \ \text{s.t.} \ -\epsilon < v_k - v_{\infty} \ \text{whenever} \ n \ge k > N_{kv} \tag{4}$$

and we want to show that

$$\forall \epsilon > 0, \exists N_{bv} > 0 \ \text{s.t.} \ |\frac{\sum_{k=1}^{n} (b_k - b_{k-1})v_k}{b_n} - v_{\infty}| < \epsilon \ \text{whenever} \ n > N_{bv} \tag{*}$$

Any errors? Can I indeed use $N_{kv}$ for both $(3)$ and $(4)$?

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If the above is correct:

1. Is $N_{kv}$ not necessarily the same as $N_{nv}$?

2. So $N = N_{kv}$?

3. It seems like we have $1 < N \le k \le n$ and $N < n$. Do we?

If $N = N_{kv}$, then I guess $N \le k$, $k \le n$ and $N < n$?

What about $N > 1$? It's not necessarily true but if it's not true, then is it that the proof is easily fixed as follows:

$$\liminf \frac{1}{b_n} \sum_{k=1}^{n} (b_k - b_{k-1})v_k \ge \liminf \frac{b_n - b_1}{b_n} (v_{\infty} - \epsilon)$$

?