TITLE: Finding $\sum\limits_{k=0}^n k$ using summation by parts
QUESTION [6 upvotes]: This is another exercise from Smoryński's Logical Number Theory; not being a mathematician, I'm a bit new to this finite difference stuff, so, please, bear with me! In a previous exercise, Smoryński asked us to prove the following formula for summation by parts:

$\sum\limits_{k=0}^n f(k)g(k) = g(n)(\sum\limits_{k=0}^n f(k)) - \sum\limits_{k=0}^{n-1} [\Delta g(k) (\sum\limits_{i=0}^k f(i))]$

Next, he wants us to apply this formula to derive $\sum\limits_{k=0}^n k = \frac{n(n+1)}{2}$. I've tried but I obviously got something wrong. First, note the obvious $\sum\limits_{k=0}^n 1 = n+1$ and $\Delta k = 0$ for $k$ constant. Thus, I got:
$\sum\limits_{k=0}^n k = \sum\limits_{k=0}^n 1 \times k = n (\sum\limits_{k=0}^n 1) - \sum\limits_{k=0}^{n-1}[\Delta k \sum\limits_{i=0}^k 1] = n(n+1) - \sum\limits_{k=0}^{n-1}0 = n(n+1) - 0 = n(n+1)$
Where did I go wrong?

REPLY [6 votes]: Let $f(k)=1$ and $g(k)=k$.  Then we have 
$$\begin{align}
S&\equiv \sum_{k=0}^{n}f(k)g(k)\\\\
&=\sum_{k=0}^{n}k\\\\
&=g(n)\sum_{k=0}^{n}f(k)-\sum_{k=0}^{n-1}\left((g_{k+1}-g_k)\sum_{i=0}^{k}f(i)\right)\\\\
&=n(n+1)-\sum_{k=0}^{n-1}\sum_{i=0}^{k}1\\\\
&=n(n+1)-\sum_{k=0}^{n-1}(k+1)\\\\
&=n(n+1)-\sum_{k=1}^{n}k\\\\
&=n(n+1)-\sum_{k=0}^{n}k\\\\
2S&=n(n+1)\\\\
S&=\frac{n(n+1)}{2}
\end{align}$$
as expected!