Are there any general rules to find
$???\leqslant \sum_{n=t}^{m}f(n)\leqslant ???$
when $m$ and $t$ $\in $ R
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1are the properties of the function (monotony,continuity,..) given? – sigmatau May 18 '13 at 11:10
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no only on general if thier is no answer so f(n) is continuity when n $in$ R – mnsh May 18 '13 at 11:12
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2Well then: no, there is no such general rule for such a general function $,f,$ . – DonAntonio May 18 '13 at 11:27
3 Answers
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There is always :
$$\sum_{n=t}^m f(n)\le (m-t+1)\max_{x\in E}\left| f(x)\right|\; \text{ where }\; E = \{t,t+1,…,m-1,m\}$$
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I think its $m-t+1$. If you sum from $1$ to $2$ you have $(2-1)+1$ elements in the sum – Belgi May 18 '13 at 11:33
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A trivial, yet sometimes useful, inequality is $$(m-t+1)\min_{i\in\{t,t+1,...,m\}}f(i)\leq\sum_{n=t}^{m}\ f(n) \leq(m-t+1)\max_{i\in\{t,t+1,...,m\}}f(i)$$
Belgi
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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 $$ \sum\nolimits_{\,n\, = \,p}^{\;q} n = \sum\limits_{p\, \le \,\,n\, < \,q} n = \sum\limits_{p\, \le \,\,n\, \le \,q - 1} n \quad \left| \matrix{ \;p \le q \hfill \cr \;p,q \in Z \hfill \cr} \right. $$
G Cab
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