Sapogov's test and the problem I have with proof Sapogov's test states that:

If $\left\{b_{k}\right\}_{k\in \mathbb N^{+}}$ is a positive monotone increasing sequence,then the series 
  $$\sum_{k=1}^{\infty}\left(1-\frac{b_{k}}{b_{k+1}}\right)$$ as well as $$\sum_{k=1}^{\infty}\left(\frac{b_{k+1}}{b_{k}}-1\right)$$ converges if the sequence $\left\{b_{k}\right\}_{k\in \mathbb N^{+}}$ is bounded, and diverges otherwise

I tried to prove this test:
Define $$a_{N}\ :=\sum_{k=1}^{N}\left(\frac{b_{k+1}}{b_{k}}-1\right)\tag{$N\in \mathbb N^{+}$}$$
And observe that $\forall k\in \mathbb N^{+}:b_k \neq0$
Then :
$$a_{N+1}-a_{N}=\sum_{k=1}^{N+1}\left(\frac{b_{k+1}}{b_{k}}-1\right)-\sum_{k=1}^{N}\left(\frac{b_{k+1}}{b_{k}}-1\right)=\frac{b_{N+2}}{b_{N+1}}-1\ge0$$
Which follows from the fact that $\left\{b_{k}\right\}$ is a positive monotone increasing sequence.
On the other hand $0\le\sum_{k=1}^{N}\left(\frac{b_{k+1}}{b_{k}}-1\right)$, so we see that $\left\{a_{N}\right\}_{N\in \mathbb N^{+}}$ is a monotone increasing bounded sequence, and hence converges, the same strategy can be used for the other series.
The problem is that I have not used the fact that $\left\{b_{k}\right\}$ is bounded, indeed I should assume such sequence is bounded and then start the proof, and for the divergence I should assume if such sequence is not bounded the the two series diverges,so where was I wrong? how can I prove the test?

Updated: 
I will show that if $\left\{b_{k}\right\}_{k\in \mathbb N^{+}}$ is a positive bounded monotone increasing sequence then both series converge:
First note that the two sequences $\left\{\zeta_{n}\right\}_{n\in \mathbb N^{+}},\left\{\rho_{n}\right\}_{n\in \mathbb N^{+}}$ defined by: 
$$\zeta_n:=\sum_{k=1}^{n}\left(1-\frac{b_{k}}{b_{k+1}}\right)\;\;\;\;\;,\;\;\;\;\rho_{n}:=\sum_{k=1}^{n}\left(\frac{b_{k+1}}{b_{k}}-1\right)$$
form a monotone increasing sequence and observe that since $\left\{b_{k}\right\}_{k\in \mathbb N^{+}}$ is bounded and strictly positive hence $\epsilon<b_{k}\le M$ for $M \in \mathbb R^{+}$ and $\epsilon>0$ , it follows:
$$\sum_{k=1}^{n}\left(1-\frac{b_{k}}{b_{k+1}}\right)\le\sum_{k=1}^{n}\left(\frac{b_{k+1}}{b_{k}}-1\right)<\frac{1}{ \epsilon}\sum_{k=1}^{n}\left(b_{k+1}-b_{k}\right)=\frac{1}{\epsilon }\left(b_{n}-b_{1}\right)\le\frac{1}{ \epsilon}\left(M-b_{1}\right)$$
$$\iff$$
$$\zeta_n\le\rho_{n}<\frac{1}{ \epsilon}\left(M-b_{1}\right)$$
Since $\left\{\zeta_{n}\right\}_{n\in \mathbb N^{+}},\left\{\rho_{n}\right\}_{n\in \mathbb N^{+}}$ are two bounded monotonic (increasing) sequences, hence they both are convergent.

Now I just don't know how to prove the divergence part.
 A: Let $c_k:=\frac{b_{k+1}}{b_k}-1$. You have already shown that $c_k$ is positive but not the potential convergence of $\sum_k c_k$, since you only gave a lower bound.
If $(b_k)$ is bounded, then (by increasingness), $(b_k)$ converges to some positive $b$ and $c_k$ is equivalent to $(b_{k+1}-b_k)/c$ and the series $\sum_k (b_{k+1}-b_k)/c$ converges. 
Suppose that $(b_k)$ is unbounded. We can find an increasing sequence of positive integers 
$(N_\ell)$ such that $b_{N_{\ell+1}-1}\geqslant 2b_{N_\ell}$. This sequence can be constructed inductively by using the definition of a sequence diverging to infinity. If $n_1,\dots,N_\ell$ are constructed, 
we know that for $n$ large enough, $b_{n+1}\geqslant 2b_{N_\ell}$. Choose such an $n$ bigger than $N_\ell$ and call it $N_{\ell+1}$.
 Then 
$$
\sum_{k=N_\ell}^{N_{\ell+1}-1}c_k=\sum_{k=N_\ell }^{N_{\ell+1}-1}\frac{b_{k+1}-b_k}{b_k}\geqslant\sum_{k=N_\ell}^{N_{\ell+1}-1}\frac{b_{k+1}-b_k}{b_{N_{\ell+1}-1}} 
=\frac{b_{N_{\ell+1}}-b_{N_\ell}}{b_{N_{\ell+1}-1}}\geqslant 1/2, $$
where in the last step, we use the fact that $b_{N_{\ell+1}\geqslant b_{N_{\ell+1}-1}}$ and the inequality $b_{N_{\ell+1}-1}\geqslant 2b_{N_\ell}$.
