Proof of Schlömilch's Generalization Schlömilch's Generalization theorem states:

Let $u(n)$ be a strictly increasing sequence of positive integers such that the ratio of successive differences is bounded;there is a positive real number $N$, for which:
$$\frac{\Delta u(n) }{\Delta u(n-1) }=\frac{u\left(n+1\right)-u\left(n\right)}{u\left(n\right)-u\left(n-1\right)}<N$$
Then provided that  $f(n)$meets the same preconditions as in Cauchy's test, the convergence of the series $\sum_{n=1}^{\infty}f\left(n\right)$ is equivalent to the convergence of:
$$\sum_{n=1}^{\infty}\Delta u(n)f\left(u\left(n\right)\right)=\sum_{n=1}^{\infty}\left(u\left(n+1\right)-u\left(n\right)\right)f\left(u\left(n\right)\right)$$
the Cauchy condensation test emerges as a special case.


I searched for a proof of this theorem, but I could not find and it seems that the theorem is not famous that much, can anybody prove this and if it's possible then gives some examples?
Updated

Given a real non-negative decreasing sequence $\left\{a_{i}\right\}_{i\in \mathbb N^{+}}$, 
Define:
$$S_{n}:=\sum_{i=1}^{n}a_{i}\;\;\;\;\;,\;\;\;\;\;\;u_i:=b^i\;\;\;\;\;,\;\;\;\;\;u_{-1}:=b^{-1}$$ Where $b \in \mathbb N_{\ge2}$ and $i\in \mathbb N$.
Clearly $u_i$ forms a sequence of strictly positive integers.
Since the terms $a_i$'s are non-negative hence:
$$S_{n+1}-S_{n}=\sum_{i=1}^{n+1}a_{i}-\sum_{i=1}^{n}a_i=a_{n+1}\ge0$$
Which follows the sequence $\left\{S_{n}\right\}_{n \in \mathbb N^{+}}$ is monotone increasing,using this fact we see that
If $n<u_{k}$ then:
$$S_{n}=\sum_{i=1}^{n}a_{i}\le\sum_{i=1}^{u_{k}}a_{i}\le a_{1}+a_{2}+a_{3}+a_{4}+...+a_{u_{k}}+...+a_{u_{k+1}-1}$$$$=\left(u_{1}-u_{0}\right)a_{u_{0}}+\left(u_{2}-u_{1}\right)a_{u_{1}}+...+\left(u_{k+1}-u_{k}\right)a_{u_{k}}=\sum_{i=0}^{k}\left(u_{i+1}-u_{i}\right)a_{u_{i}}=\zeta_k$$
$$\iff$$
$$\bbox[5px,border:2px solid #C0A000]{S_{n}\le\zeta_k}\tag{I}$$
If $n>u_{k}$ then:
$$S_{n}=\sum_{i=1}^{n}a_{i}\ge\sum_{i=1}^{u_{k}}a_{i}=a_{1}+a_{2}+a_{3}+a_{4}+...+a_{u_{k-1}+1}+...+a_{u_{k}}$$$$\ge\left(u_{0}-u_{-1}\right)a_{u_{0}}+\left(u_{1}-u_{0}\right)a_{u_{1}}+...+\left(u_{k}-u_{k-1}\right)a_{u_{k}}=\sum_{i=0}^{k}\left(u_{i}-u_{i-1}\right)a_{u_{i}}=\frac{\zeta_k}{u_1}$$
$$\iff$$
$$\bbox[5px,border:2px solid #C0A000]{u_1S_{n}\ge\zeta_k}\tag{II}$$
(to be more precise $(\text{II})$ holds for any $M\ge u_{1}\ge 2$)
Combining $(\text{I})$ and $(\text{II})$ follows:
$S_{n}$ converges if and only if $\zeta_k$ converges and diverges if and only if $\zeta_k$ diverges.
Setting $u_i\mapsto2^i$ gives Cauchy condensation test .

The problem is that I have not used the facts that $u_i>u_{i-1}$ (I know this is true) and for a positive $M$ : $$\frac{u_{k+1}-u_{k}}{u_{k}-u_{k-1}}\le M$$
This is true if and only if $M\ge u_{1}\ge2$
(I again know that this is true).
So what is wrong with my proof? However I know that these two are true, but Since I've proved the theorem without using these two facts I'm not sure if the proof is right.
I think as long as $S_{n}\le\zeta_k\le MS_{n}$ holds for $M\ge u_{1}$ we are able to use the test (moreover since $u_i$ forms a strictly increasing sequence then if $M=u_{i}$ for natural $i\ge1$ then again we can use the test, that's what I think Wikipedia says).

Updated:
From page 122 of Theory And Application Of Infinite Series by Knopp,Konrad, I know that my definiton of $u_i$ is a special case, however I still don't know why do we need 
 A: Here's the complete proof if my memory serves me correctly, I always found the proof in Knopp a bit unpleasant to read so I tried to make it clearer.
We have by assumption that $f(n)$ is decreasing and $u_n$ is non-negative strictly increasing satisfying $\Delta u_n \leq C \Delta u_{n-1}$ for a $C > 0$, call the partial sums of the series $s_m = \sum_{n=0}^{m} f(n)$ and the partial sums of the condensed series by $t_m = \sum_{n=0}^{m} \Delta u_{n} f\left(u_{n}\right)$
Define $\displaystyle A=\sum_{n \geq 0: f(n)<f\left(u_{0}\right)} f(n) \quad$ when the index $m$ is such that $m < u_k$ we have the inequality
$$\begin{aligned}
s_{m} \leq s_{u_{k}} & =  A+\Big(f(u_{0})+f(u_{0}+1)+ \dots f\left(u_{1}-1\right)\Big)\\ & +\ldots+\mkern-1mu\Big(f\left(u_{k}\right)+f\left(u_{k}+1\right)+\ldots+f\left(u_{k+1}-1\right)\Big) \\
& \leq A+\mkern2mu\left(u_{1}-u_{0}\right) f\left(u_{0}\right)+\ldots+\left(u_{k+1}-u_{k}\right) f\left(u_{k}\right) \\[0.25cm] \text {since } f(n) \text { decreasing } \\{}\\
&=A+t_{k}
\end{aligned}$$
The first line is breaking up the sum of the first $u_k$ many terms by $A$ and whatever is left over, then we group up the terms in the large brackets and estimate each $f$ with the largest choice in the large bracket.
When the index $m$ is such that $m>u_{k}$
$$
\begin{aligned}
s_{m} \geq s_{u_{k}} & \geq\Big(f\left(u_{0}+1\right)+f\left(u_{0}+2\right)+\ldots+f\left(u_{1}\right)\Big)+\ldots+\Big(f\left(u_{k-1}+1\right)+\ldots+f\left(u_{k}\right)\Big) \\
& \geq\left(u_{1}-u_{0}\right) f\left(u_{1}\right)+\ldots+\left(u_{k}-u_{k-1}\right) f\left(u_{k}\right)\\
& = \sum_{j=1}^k\Delta u_{j-1}f(u_j)\\
& \geq \sum_{j=1}^k\frac{1}{C}\Delta u_{j}f(u_j)
\end{aligned}
$$
The first and second line use the same trick as before, but this time we estimate each $f$ with the lowest choice in the large brackets, and the final inequality uses the assumption on $u_n$.
The last inequality leads to
$$
C s_{m} \geq t_{k}-t_{0}
$$
So we have shown that $t_k$ and $s_k$ always bound one another above and below (we don't worry about the conditions $m > u_k$ and so on since we can make either one as big as we need in the infinite sum), thus the convergence and divergence of one are determined by the other.
The special case $u_n = 2^n = \Delta u_n\;, C=1$ is the Cauchy condensation test.
