questions about Rudin's summation by parts rudin's summation by parts

3.41  Theorem Given two sequences $\left\{a_n\right\}$,$\left\{b_n\right\}$,put
$$\begin{align*}A_n=\sum _{k=0}^n a_k~~~(1)\end{align*}$$
if $n\geq 0$; put $A_{-1}=0$. Then, if $0\leq p\leq q$, we have
$$\begin{align*}\sum _{n=p}^q a_nb_n=\sum _{n=p}^{q-1} A_n\left(b_n-b_{n+1}\right)+\color{blue}{A_qb_q}\color{red}{-}\color{blue}{A_{p-1}b_p}.~~~(2)\end{align*}$$
wiki's summation by parts
My derivation:

blue color highlight the changed item.

Questions

question1:

The blue part of (2) is a little hard to memorize before I've done the calculation.
What's the relation between (2) and the Integration by parts which I think is easier to memorize (without those indexes).
$\int f dg=f g-\int f dg$
or
$\int _a^bF(x)g(x)dx=F(b)G(b)-F(a)G(a)-\int _a^bf(x)G(x)dx$
question2:

What's the guidelines to derive the (2)?
As you see the wiki's edition, there are also several versions.
Although I get (2), I did it by cheating, i.e. I'm doing proof, and I checked (2) in my process of derivation.
Of course, the proof in text, is much easier:
$$\begin{align*}\sum _{n=p}^q a_nb_n=\sum _{n=p}^q \left(A_n-A_{n-1}\right)b_n=\sum _{n=p}^q A_nb_n-\sum _{n=p-1}^{q-1} A_nb_{n+1}~~~(4)\end{align*}$$
,and the last expression on the right is clearly equal to the right side of (2)
question3

$A_n$is summation from $0$ to $n$, how does it look reasonable when $p=5$?
 A: Define
$$
A_n=\sum_{k=1}^na_k\quad\text{and}\quad F(x)=\int_a^xf(t)\,\mathrm{d}t
$$
Question 1:
Consider summation by parts
$$
\sum_{k=m}^na_kb_k
=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k(b_{k+1}-b_k)
$$
vs integration by parts
$$
\int_a^bf(x)g(x)\,\mathrm{d}x=F(b)g(b)-F(a)g(a)-\int_a^bF(x)g'(x)\,\mathrm{d}x
$$
The parallel seems pretty clear.

Question 2:
I'm not sure what you mean by guidelines to derive summation by parts. Basically, summation by parts is just a change of the index of summation.
$$
\begin{align}
\sum_{k=m}^na_kb_k
&=\sum_{k=m}^n(A_k-A_{k-1})b_k\\
&=\sum_{k=m}^nA_kb_k-\sum_{k=m}^nA_{k-1}b_k\\
&=\sum_{k=m}^nA_kb_k-\sum_{k=m-1}^{n-1}A_kb_{k+1}\\
&=A_nb_n+\sum_{k=m}^{n-1}A_kb_k-A_{m-1}b_m-\sum_{k=m}^{n-1}A_kb_{k+1}\\
&=A_nb_n-A_{m-1}b_m+\sum_{k=m}^{n-1}A_k(b_k-b_{k+1})
\end{align}
$$

Question 3:
$$
\sum_{k=5}^na_kb_k
=A_nb_n-A_4b_5-\sum_{k=5}^{n-1}A_k(b_{k+1}-b_k)
$$
It looks reasonable to me. If you're worried about where the summation starts, that only alters $A_n$ by a constant. Consider how much a constant added to $A_n$ changes the value of the summation by parts formula:
$$
\begin{align}
&(A_n+C)b_n-(A_{m-1}+C)b_m-\sum_{k=m}^{n-1}(A_k+C)(b_{k+1}-b_k)\\
&=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k(b_{k+1}-b_k)\\
&+C\left(b_n-b_m-\sum_{k=m}^{n-1}(b_{k+1}-b_k)\right)\\
&=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k(b_{k+1}-b_k)\\
&+C(b_n-b_m-(b_n-b_m))\\
&=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k(b_{k+1}-b_k)
\end{align}
$$
$C$ has no effect on the final value of the formula.
