Bounding the sum by an integral

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Bounding the sum by an integral

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[note article incomplete]More examples are needed and more complicated situations should be analyzed. [/note]

[QUICK DESCRIPTION]
Suppose we want to estimate a sum

[math] \sum_{a<n<b} a_n [/math]

where the sequence $\{a_n\}$ is a monotone sequence of non negative real numbers and $a<b$ are real numbers. Later we can let for example $b\rightarrow +\infty$ and likewise $a\rightarrow -\infty$ but let us keep $a,b$ finite for the moment. In this case is a standard tactic to look for a real function $f$ with the same kind of monotonicity as the sequence $\{a_n\}$, such that $f(n)=a_n$ for all the $n$ we are interested in. Then we have

[math] \sum _{a<n<b} a_n \simeq \int_a ^b f(x) dx . [/math]

Usually the choice of the function should be obvious by looking at the sequence $\{a_n\}$. If for example the sequence $\{a_n\}$ is explicitly given then the first obvious choice would be to replace the discrete parameter $n$ with a continuous variable $x$ and look at the resulting function $f(x)$.

[PREREQUISITES]
calculus

[EXAMPLE]
Let us look at the partial sums of the harmonic series

[math] \sum_{k=1} ^n \frac{1}{k}.[/math]

The sequence $a_k=\frac{1}{k}, k=1,2,\ldots$ is a strictly decreasing sequence of positive numbers. Taking $f(x)=\frac{1}{x}$ we recover the well known estimate

[math] \sum_{k=1} ^n \frac{1}{k}\simeq \int_1 ^n \frac{dx}{x}= \log n[/math]

[EXAMPLE]

One can use the same technique to prove that for a positive real number $p$, the '''over-harmonic series'''

[math]\sum_{k=1} ^\infty \frac{1}{n^p}, [/math]

converge exactly when $p>1$ and we have the estimate

[math]\sum_{k=1} ^\infty \frac{1}{n^p}\simeq  \frac{1}{p-1},[/math]

whenever $p>1$.
[GENERAL DISCUSSION]

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Bounding the sum by an integral

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