In our lecture we have two different ways to define summation by parts:
$$1.)~~\sum\limits_{k=0}^na_k(b_{k+1}-b_k)=a_nb_{n+1}-a_0b_0 -\sum\limits_{k=1}^nb_k(a_{k}-a_{k-1})\\ 2.)~~\sum\limits_{k=0}^na_kb_k=a_nB_n+\sum\limits_{k=0}^{n-1}B_k(a_k-a_{k+1}),$$ where $ B_k:=\sum\limits_{i=0}^kb_i$ and $(a_k), (b_k)$ are two real valued seqeunces.
Show that both statements are equivalent.
My approach:
$1.)\implies 2.)$:
We define the series $S_n$ as follows: $$S_0:=0,\\ S_k:=B_{k-1}\\\implies S_{k+1}-S_{k}=b_k.$$ We use this definition and plug $(S_{k+1}-S_{k})$ into $2.)$: $$ \sum\limits_{k=0}^na_kb_k = \sum\limits_{k=0}^na_k(S_{k+1}-S_{k})\underset{\text{we apply }1.)}{=}a_nS_{n+1}-a_0S_0-\sum\limits_{k=1}^n S_k(a_k-a_{k-1})\\=a_nB_n-a_0\cdot 0+\sum\limits_{k=1}^n B_{k-1}(a_{k-1}-a_k)=a_nB_n+\sum\limits_{k=0}^{n-1} B_{k}(a_{k}-a_{k+1}). $$ $2.)\implies 1.)$:
We define $d_k:=b_{k+1}-b_k$ and $D_k:=\sum\limits_{i=0}^{k}d_i=b_{k+1}-b_0$ and plug it into $1.)$: $$ \sum\limits_{k=0}^na_k(b_{k+1}-b_k)= \sum\limits_{k=0}^na_kd_k\underset{\text{we apply }2.)}{=}a_nD_n+\sum\limits_{k=0}^{n-1}D_k(a_k-a_{k+1})\\=a_n(b_{n+1}-b_0)+\sum\limits_{k=0}^{n-1}(b_{k+1}-b_0)(a_k-a_{k+1})\\ =a_nb_{n+1}-a_nb_0+\sum\limits_{k=0}^{n-1}b_{k+1}(a_k-a_{k+1})-\sum\limits_{k=0}^{n-1}b_0(a_k-a_{k+1})\\=a_nb_{n+1}-a_nb_0-\sum\limits_{k=1}^{n}b_{k}(a_k-a_{k-1})-b_0(a_0-a_{n})\\ =a_nb_{n+1}-a_0b_0-\sum\limits_{k=1}^{n}b_{k}(a_k-a_{k-1}). $$
My tutor told me that this is wrong and I can't do it this way. Did I mix up the indices or is this approach in general wrong? Any comments or hints are appreciated :)