question about summation? Are there any general rules to find 
$???\leqslant \sum_{n=t}^{m}f(n)\leqslant ???$
when $m$ and $t$ $\in $ R
 A: There is always :
$$\sum_{n=t}^m f(n)\le (m-t+1)\max_{x\in E}\left| f(x)\right|\; \text{ where }\; E = \{t,t+1,…,m-1,m\}$$
A: A trivial, yet sometimes useful, inequality is $$(m-t+1)\min_{i\in\{t,t+1,...,m\}}f(i)\leq\sum_{n=t}^{m}\ f(n) \leq(m-t+1)\max_{i\in\{t,t+1,...,m\}}f(i)$$
A: Note that the expression
$$
\sum\nolimits_{\,n\, = \,a}^{\;b} {f(n)} 
$$
has a very precise and rigourous definition, valid for $a$ and $b$ real or even complex,
as far as $f(n)$ is defined also for $n$ real or complex.
The definition is given by the Indefinite Sum, that is
$$
f(x) = \Delta _x F(x) = F(x + 1) - F(x)\quad  \Leftrightarrow \quad \sum\nolimits_{\,n\, = \,a}^{\;b} {f(n)}  = F(b) - F(a)
$$
Example:
$$
f(n) = n\quad  \to \quad f(x) = x\quad  \to \quad \Delta _x \left( {{{x\left( {x - 1} \right)} \over 2}} \right)
$$
so that
$$
\sum\nolimits_{\,n\, = \,a}^{\;b} n  = {{b\left( {b - 1} \right) - a\left( {a - 1} \right)} \over 2} = {{\left( {b - a} \right)\left( {b + a - 1} \right)} \over 2}
$$
Finally note that for integral bounds we have
$$
\sum\nolimits_{\,n\, = \,p}^{\;q} n  = \sum\limits_{p\, \le \,\,n\, < \,q} n  = \sum\limits_{p\, \le \,\,n\, \le \,q - 1} n \quad \left| \matrix{
  \;p \le q \hfill \cr 
  \;p,q \in Z \hfill \cr}  \right.
$$
