How to Prove $ \sum_{k=1}^{n} \Theta(f(k)) = \Theta(\sum_{k=1}^{n} f(k))$ In the book Introduction to Algorithm by Thomas H. Cormen et al., this is introduced:
$$\sum_{k=1}^{n} \Theta\big(f(k)\big) = \Theta\big(\sum_{k=1}^{n} f(k)\big)$$
I need help with a better explanation on how to prove this. The quote below is from the book directly. I do understand the first summation, I just don't understand the Asymptotic Big Theta summation.

Linearity
For any real number $c$ and any finite sequence $a_1,a_2,\cdots,a_n$ and $b_1,b_2,\cdots,b_n$,
$$\sum^{n}_{k \, = \, 1}(ca_k + b_k) = c \sum^{n}_{k \, = \, 1} a_l + \sum^{n}_{k \, = \, 1} b_k$$
The linearity property also applies to infinite convergent series.
We can explit the linearity property to manipulate summations incorporating asymptotic notation. For example,
$$\sum_{k=1}^{n} \Theta\big(f(k)\big) = \Theta\big(\sum_{k=1}^{n} f(k)\big)$$
In this equation, the $\Theta$-notation on the left-hand side applies to the variable $k$, but on the right-hand side, applies to $n$. We can also apply such manipulations to infinite convergent series.

 A: Let me write my thoughts, taken from one of my unfinished article, in place of the answer, in order to attract more attention of our community and, as a result, more criticism. I ask all possible down voters to express their views explicitely, given that this will only serve the benefit of our site, its core spirit of being a place of quality resources, and, also, for all of its current and future users. I personally would be very grateful.
To emphasize the brough thought, I give all the necessary definitions below.
Coming back to question on left hand there is
$$\sum\limits_{k=1}^{n}\Theta (f(k) )\quad (1)$$
and is written, that "$\Theta$ notation on the left hand side is applied to the variable $k$ ". But if we understand this words, so that we have $\Theta (f(k) ),k \to \infty$, look definition below, then $k$ is bound variable. So it can be replaced with any other symbol, unequal used symbols, without changing the meaning. This lost its connection with outer sum.
Let's  write $$\sum\limits_{k=1}^{n}\lim\limits_{k\to\infty} f(k) \quad (2)$$
what sense can be given to this formula ? Even if we consider the variable $k$ in the sum and in the $\lim$ as "independent", as is sometimes done when integrating with a variable upper limit, then for $(2)$ we get simply $n \cdot\lim\limits_{k\to\infty} f(k)$. But even this have not big sense for $(1)$, because as $n$ is constant with respect to $\Theta$'s variable $k$, then we have $n \cdot \Theta (f(k) ) = \Theta (f(k) ), k\to\infty$.
Thus, to understand $(1)$ it must be given a meaning different from the classical definitions, given below, and generally accepted in mathematics. In this case, this should be explicitly wrote, although, in this case, there must be a reason why to abandon generally understandable definitions and mislead readers with ordinary mathematical education. I gave one of the possible correct visions of the asked formula above in the comment.
Finally, once again, I ask all possible down voters to express their criticism explicitly in the comments. Prephasing the famos “Strike, if you will, but listen.”, let me write
$$\text{Strike, if you will, but argue.}$$
Definitions.
Having a monoid with a binary operation on it, in algebra, for finite index family, is inductively defined, so called, composite sum
\begin{equation}\label{1}
\sum\limits_{k=1}^1 x_k = x_1, \\ 
\sum\limits_{k=1}^{n+1} x_k = \left(\sum\limits_{k=1}^{n} x_k \right) + x_{n+1} 
\end{equation}
It can be shown, that in monoid we can put brackets in any manner, which gives possibility not to set brackets at all and write composite sum for several summands as
\begin{equation}\label{3}
\sum\limits_{k=1}^{n} x_k = x_1+ x_2+ \cdots + x_n
\end{equation}
$\boldsymbol{\sum}$ is known, also, as variable-binding operator. The symbol $k$, denoting the variable index, is bound variable here and can be replaced with any other symbol unequal with $\boldsymbol{n}$, which is free variable, and without changing the meaning of sum.
Now let's take definition of big-$\Theta$ for non negative case
$$\Theta(f) = \left\lbrace g \colon \exists C_1>0,\exists C_2>0,\exists N \in \mathbb{N}, \forall n \in\mathbb{N} \land n > N, C_1 \cdot f(n) \leqslant g(n)\leqslant C_2 \cdot f(n)  \right\rbrace  $$
To emphasize variable with respect to which $\Theta$ is applied, because $f$ can be multivariable, often we write $\Theta(f(n)), n\to\infty$. As we see definition contains only one free variable $f$ and constant $\mathbb{N}$. All other variables - $g, C_1,C_2 , N, n$ - are bound, dummy, variables, so can be replaced with any other symbol unequal to $f$ and each other without changing the meaning.
