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
$$\begin{align*}\sum _{n=p}^q a_nb_n=\sum _{n=p}^{q-1} A_n\left(b_n-b_{n+1}\right)+\color{blue}{A_qb_q}\color{red}{-}\color{blue}{A_{p-1}b_p}.~~~(2)\end{align*}$$
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 $p=5$?
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.